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Section 13.2 Quadratic Graphs and Vertex Form

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In this section, we will explore quadratic functions using graphing technology and learn the vertex and factored forms of a quadratic function’s formula. We will also see how parabola graphs can be shifted.
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Figure 13.2.1. Alternative Video Lessons
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Subsection 13.2.1 Exploring Quadratic Functions with Graphing Technology

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Graphing technology is very important and useful for applications and for finding points quickly. Let’s explore some quadratic functions with graphing technology.
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Example 13.2.2.

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Use technology to graph and make a table of the quadratic function f defined by f(x)=2x2+4x−3 and find each of the key points or features.
  1. Find the vertex.
  2. Find the vertical intercept (i.e. y-intercept).
  3. Find the horizontal or (i.e. x-intercept(s)).
  4. Find f(−2).
  5. Solve f(x)=3 using the graph.
  6. Solve f(x)≤3 using the graph.
  7. State the domain and range of the function.
Explanation.
The specifics of how to use any one particular technology tool vary. Whether you use an app, a physical calculator, or something else, a table and graph should look like:
x f(x)
−2 −3
−1 −5
0 −3
1 3
2 13
the graph of the parabola f(x)=2x^2+4x-3, that opens upward, passing through the points listed in the table
Additional features of your technology tool can enhance the graph to help answer these questions. You may be able to make the graph appear like:
the previous graph with all of the key points labeled as described below
  1. The vertex is (−1,−5).
  2. The vertical intercept is (0,−3).
  3. The horizontal intercepts are approximately (−2.6,0) and (0.6,0).
  4. When x=−2, y=−3, so f(−2)=−3.
  5. The solutions to f(x)=3 are the x-values where y=3. We graph the horizontal line y=3 and find the x-values where the graphs intersect. The solution set is {−3,1}.
  6. The solutions are all of the x-values where the function’s graph is below (or touching) the line y=3. The interval is [−3,1].
  7. The domain is (−∞,∞) and the range is [−5,∞).
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Now we will look at an application with graphing technology and put the points of interest in context.
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Example 13.2.3.

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A reduced-gravity aircraft
 1 
en.wikipedia.org/wiki/Reduced-gravity_aircraft
is a fixed-wing airplane that astronauts use for training. The airplane flies up and then down in a parabolic path to simulate the feeling of weightlessness. In one training flight, the pilot will fly 40 to 60 parabolic maneuvers.
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For the first parabolic maneuver, the altitude of the plane, in feet, at time t, in seconds since the maneuver began, is given by H(t)=−16t2+400t+30500.
  1. Determine the starting altitude of the plane for the first maneuver.
  2. What is the altitude of the plane 10 seconds into the maneuver?
  3. Determine the maximum altitude of the plane and how long it takes to reach that altitude.
  4. The zero-gravity effect is experienced when the plane begins the parabolic path until it gets back down to 30,500 feet. Write an inequality to express this and solve it using the graph. Write the times of the zero-gravity effect as an interval and determine how long the astronauts experience weightlessness during each cycle.
  5. Use technology to make a table for H with t-values from 0 to 25 seconds. Use an increment of 5 seconds and then use the table to solve H(t)=32100.
  6. State the domain and range for this context.
Explanation.
We can answer the questions based on the information in the graph.
a graph of the parabola y=-16t^2+400t+30500 and the horizontal line y=30500; the parabola opens downward and has a maximum at (12.5, 33000)
Figure 13.2.4. Graph of H(t)=−16t2+400t+30500 with y=30500
  1. The starting altitude can be read from the vertical intercept, which is (0,30500). The feeling of weightlessness begins at 30,500 feet.
  2. After 10 seconds, the altitude of the plane is 32,900 feet.
  3. For the maximum altitude of the plane we look at the vertex, which is approximately (12.5,33000). This tells us that after 12.5 seconds the plane will be at its maximum altitude of 33,000 feet.
  4. We can write an inequality to describe when the plane is at or above 30,500 feet and solve it graphically.
    H(t)≥30500−16t2+400t+30500≥30500.
    We graph the line y=30500 and find the points of intersection with the parabola. The astronauts experience weightlessness from 0 seconds to 25 seconds into the maneuver, or [0,25] seconds. They experience weightlessness for 25 seconds in each cycle.
  5. To solve H(t)=32100 using the table, we look for where the H-values are equal to 32100.
    t 0 5 10 15 20 25
    H(t) 30500 32100 32900 32900 32100 30500
    There are two solutions, 5 seconds and 20 seconds. The solution set is {5,20}.
  6. When we use technology we see the entire function but in this context the plane is only on a parabolic path from t=0 to t=25 seconds. So the domain is [0,25], and the range is the set of corresponding y-values which is [30500,33000] feet.
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Let’s look at the remote-controlled airplane dive from Example 9.3.18. This time we will use technology to answer the questions.
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Example 13.2.5.

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Maia has a remote-controlled airplane and she is going to do a stunt dive where the plane dives toward the ground and back up along a parabolic path. The altitude or height of the plane is given by the function H where H(t)=0.7t2−23t+200, for 0≤t≤30. The height is measured in feet and the time, t, is measured in seconds since the stunt began.
  1. Determine the starting height of the plane as the dive begins.
  2. Determine the height of the plane after 5 seconds.
  3. Will the plane hit the ground, and if so, at what time?
  4. If the plane does not hit the ground, what is the closest it gets to the ground, and at what time?
  5. At what time(s) will the plane have an altitude of 50 feet?
  6. State the domain and the range of the function (in context).
Explanation.
We have graphed the function and we will find the key information and put it in context.
a graph of the quadratic function with all of the key features labeled that are described below
Figure 13.2.6. Graph of H(t)=0.7t2−23t+200
  1. The starting altitude can be read from the vertical intercept, which is (0,200). When the stunt begins, the plane has a altitude of 200 feet.
  2. When x=5, the y-value is 102.5. So H(5)=102.5. This means that after 5 seconds, the plane is 102.5 feet above the ground.
  3. From the graph we can see that the parabola does not touch or cross the x-axis, which represents the ground. This means the plane does not hit the ground and there are no real solutions to the equation H(t)=0.
  4. The lowest point is the vertex, which is approximately (16.43,11.07). The minimum altitude of the plane is about 11 feet, which occurs after about 16.4 seconds.
  5. We graph the horizontal line y=50 and look for the points of intersection. The plane will be 50 feet above the ground about 9 seconds after the plane begins the stunt, and again at about 24 seconds.
  6. The domain for this function is given in the problem statement because only part of the parabola represents the path of the plane. The domain is [0,30]. For the range we look at the possible altitudes of the plane and see that it is [11.07…,200]. The plane is doing this stunt from 0 to 30 seconds and its height ranges from about 11 to 200 feet above the ground.
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Subsection 13.2.2 The Vertex Form of a Parabola

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We have learned the standard form of a quadratic function’s formula, which is f(x)=ax2+bx+c. In this subsection, we will learn another form called the “vertex form”.
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Using graphing technology, consider the graphs of f(x)=x2−6x+7 and g(x)=(x−3)2−2 on the same axes.
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We see only one parabola because these are two different forms of the same function. Indeed, if we convert g(x) into standard form:
g(x)=(x−3)2−2g(x)=x2−6x+9−2g(x)=x2−6x+7
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it is clear that f and g are the same function.
the graphs of the two parabolas overlap each other completely
Figure 13.2.7. Graph of f(x)=x2−6x+7 and g(x)=(x−3)2−2
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The formula given for g is said to be in “vertex form” because it allows us to read the vertex without doing any calculations. The vertex of the parabola is (3,−2). We can see those numbers in g(x)=(x−3)2−2. The x-value is the solution to (x−3)=0, and the y-value is the constant added at the end.
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Here are the graphs of three more functions with formulas in vertex form. Compare each function with the vertex of its graph.
the graph of a parabola that opens upward and has a vertex at (2,1)
Figure 13.2.8. r(x)=(x−2)2+1
the graph of a wide parabola that opens downward and has a vertex at (-1,3)
Figure 13.2.9. s(x)=−14(x+1)2+3
the graph of a narrow parabola that opens upward and has a vertex at (-3,-3.5)
Figure 13.2.10. t(x)=4(x+3)2−3.5
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Notice that the x-coordinate of the vertex has the opposite sign as the value in the function formula. On the other hand, the y-coordinate of the vertex has the same sign as the value in the function formula. Let’s look at an example to understand why. We will evaluate r(2).
r(2)=(2−2)2+1=1
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The x-value is the solution to (x−2)=0, which is positive 2. When we substitute 2 for x we get the value y=1. Note that these coordinates create the vertex at (2,1). Now we can define the vertex form of a quadratic function.
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Checkpoint 13.2.12.

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Find the vertex of each quadratic function.
  1. r(x)=−2(x+4)2+10
  2. s(x)=5(x−1)2+2
  3. t(x)=(x−10)2−5
  4. u(x)=3(x+7)2−13
Explanation.
  1. The vertex of r(x)=−2(x+4)2+10 is (−4,10).
  2. The vertex of s(x)=5(x−1)2+2 is (1,2).
  3. The vertex of t(x)=(x−10)2−5 is (10,−5).
  4. The vertex of u(x)=3(x+7)2−13 is (−7,−13).
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Now let’s do the reverse. When given the vertex and the value of a, we can write the function in vertex form.
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Example 13.2.13.

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Write a formula for the quadratic function f with the given vertex and value of a.
  1. Vertex (−2,8), a=1
  2. Vertex (4,−9), a=−4
  3. Vertex (−3,−1), a=2
  4. Vertex (5,12), a=−3
Explanation.
  1. The vertex form is f(x)=(x+2)2+8.
  2. The vertex form is f(x)=−4(x−4)2−9.
  3. The vertex form is f(x)=2(x+3)2−1.
  4. The vertex form is f(x)=−3(x−5)2+12.
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Once we read the vertex we can also state the domain and range. All quadratic functions have a domain of (−∞,∞) because we can put in any value to a quadratic function. The range, however, depends on the y-value of the vertex and whether the parabola opens upward or downward. When we have a quadratic function in vertex form we can read the range from the formula. Let’s look at the graph of f, where f(x)=2(x−3)2−5, as an example.
the graph of a parabola that opens upward with a vertex of (3,-5)
Figure 13.2.14. The graph of f(x)=2(x−3)2−5
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The domain is (−∞,∞). The graph of f opens upward (which we know because a=2 is positive) so the vertex is the minimum point. The y-value of −5 is the minimum. The range is [−5,∞).
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Example 13.2.15.

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Identify the domain and range of g, where g(x)=−3(x+1)2+6.
Explanation.
the graph of a parabola that opens downward with a vertex of (-1,6)
Figure 13.2.16. g(x)=−3(x+1)2+6
The domain is (−∞,∞). The graph of g opens downward (which we know because a=−3 is negative) so the vertex is the maximum point. The y-value of 6 is the maximum. The range is (−∞,6].
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Checkpoint 13.2.17.

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Identify the domain and range of each quadratic function.
  1. w(x)=−3(x+10)2−11
    The domain is and the range is .
  2. u(x)=4(x−7)2+20
    The domain is and the range is .
  3. y(x)=−(x−1)2
    The domain is and the range is .
  4. z(x)=3(x+9)2−4
    The domain is and the range is .
Explanation.
  1. The domain of w is (−∞,∞). The parabola opens downward so the range is (−∞,−11].
  2. The domain of u is (−∞,∞). The parabola opens upward so the range is [20,∞).
  3. The domain of y is (−∞,∞). The parabola opens downward so the range is (−∞,0].
  4. The domain of z is (−∞,∞). The parabola opens upward so the range is [−4,∞).
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Subsection 13.2.3 Horizontal and Vertical Shifts

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Let f(x)=x2 and g(x)=(x−4)2+1. The graph of y=f(x) has its vertex at the point (0,0). Now we will compare this with the graph of y=g(x) on the same axes.
the graph of two identical parabolas that open upward; one has a vertex at (0,0) and the other one is shifted to the right by 4 units and up by one unit so it’s vertex is at (4,1)
Figure 13.2.18. The graph of f and g
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Both graphs open upward and have the same shape. Notice that the graph of g is the same as the graph of f but is shifted to the right by 4 units and up by 1 units because its vertex is (4,1).
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Let’s look at another graph. Let h(x)=−x2 and let j(x)=−(x+3)2+4.
the graph of two identical parabolas that open downward; one has a vertex at (0,0) and the other is shifted to the left by 3 units and up by 4 units to have a vertex at (-3,4)
Figure 13.2.19. The graph of h and j
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Both parabolas open downward and have the same shape. The graph of j is the same as the graph of h but it has been shifted to the left by 3 units and up by 4 units making its vertex (−3,4).
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To summarize, when a quadratic function is written in vertex form, the h-value is the horizontal shift of its graph from the graph of y=ax2 and the k-value is the vertical shift of its graph from the graph of y=ax2.
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Example 13.2.20.

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Identify the horizontal and vertical shifts compared with y=x2.
  1. m(x)=(x+7)2+3
  2. n(x)=(x−1)2+6
  3. o(x)=(x−5)2−1
  4. p(x)=(x+3)2−11
Explanation.
  1. The graph of y=m(x) has vertex at (−7,3). Therefore the graph is the same as y=x2 shifted to the left 7 units and up 3 units.
  2. The graph of y=n(x) has vertex at (1,6). Therefore the graph is the same as y=x2 shifted to the right 1 unit and up 6 units.
  3. The graph of y=o(x) has vertex at (5,−1). Therefore the graph is the same as y=x2 shifted to the right 5 units and down 1 unit.
  4. The graph of y=p(x) has vertex at (−3,−11). Therefore the graph is the same as y=x2 shifted to the left 3 units and down 11 units.
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Subsection 13.2.4 The Factored Form of a Parabola

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There is another form of a quadratic function’s formula, called “factored form”, which we will explore next. Let’s consider the two functions q(x)=−x2+3x+4 and s(x)=−(x−4)(x+1). Using graphing technology, we will graph y=q(x) and y=s(x) on the same axes.
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These graphs coincide because the functions are actually the same. We can tell by multiplying out the formula for g to get back to the formula for f.
g(x)=−(x−4)(x+1)g(x)=−(x2−3x−4)g(x)=−x2+3x+4
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Now we can see that f and g are really the same function.
the graph of two identical parabolas that open downward and overlap each other completely
Figure 13.2.21. Graph of q and s
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Factored form is very useful because we can read the x-intercepts directly from the function, which in this case are (4,0) and (−1,0). We find these by looking for the values that make the factors equal to 0, so the x-values have the opposite signs as are shown in the formula. To demonstrate this, we will find the roots by solving g(x)=0.
g(x)=−(x−4)(x+1)0=−(x−4)(x+1)
x−4=0orx+1=0x=4orx=−1
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This shows us that the x-intercepts are (4,0) and (−1,0).
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The x-values of the x-intercepts are also called zeros or roots. The zeros or roots of the function g are −1 and 4.
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Checkpoint 13.2.23.

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Write the horizontal intercepts of each function.
  1. t(x)=−(x+2)(x−4)
  2. u(x)=6(x−7)(x−5)
  3. v(x)=−2(x+1)(x+4)
  4. w(x)=10(x−8)(x+3)
Explanation.
  1. The horizontal intercepts of t are (−2,0) and (4,0).
  2. The horizontal intercepts of u are (7,0) and (5,0).
  3. The horizontal intercepts of v are (−1,0) and (−4,0).
  4. The horizontal intercepts of w are (8,0) and (−3,0).
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Let’s summarize the three forms of a quadratic function’s formula:
Standard Form
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f(x)=ax2+bx+c, with y-intercept (0,c).
Vertex Form
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f(x)=a(x−h)2+k, with vertex (h,k).
Factored Form
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f(x)=a(x−r)(x−s), with x-intercepts (r,0) and (s,0).
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Reading Questions 13.2.5 Reading Questions

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1.

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With the vertex form of a quadratic function, the formula shows you a point on the graph (without having to do any calculation). What is the name of that point?
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2.

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With the standard form of a quadratic function, the formula shows you a point on the graph (without having to do any calculation). What is the name of that point?
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3.

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What makes the vertex form of a quadratic function nicer for graphing compared to standard form?
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4.

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If a fellow student attempted to graph the equation y=(x−4)2+6 and put the vertex at (-4,6), how would you explain to them that they have made an error?
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Exercises 13.2.6 Exercises

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Review and Warmup.

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1.
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Multiply the polynomials.
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(r+5)(r−9)=
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2.
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Multiply the polynomials.
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(t+1)(t−4)=
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3.
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Multiply the polynomials.
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(8t−10)(t+5)=
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4.
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Multiply the polynomials.
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(5x−6)(x+7)=
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5.
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Factor the given polynomial.
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x2+10x+9=
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6.
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Factor the given polynomial.
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y2+10y+21=
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7.
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Factor the given polynomial.
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2y2+20y+18=
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8.
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Factor the given polynomial.
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9y2+27y+18=
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9.
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For the interval expressed in the number line, write it using set-builder notation and interval notation.
Graph of an interval on a number line. There is a left bracket at 1 and everything right of that is shaded. To the right of the axis, there is a letter q.
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10.
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For the interval expressed in the number line, write it using set-builder notation and interval notation.
Graph of an interval on a number line. There is a right bracket at 2 and everything left of that is shaded. To the right of the axis, there is a letter h.
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11.
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For the interval expressed in the number line, write it using set-builder notation and interval notation.
Graph of an interval on a number line. There is a left parenthesis at 3 and everything right of that is shaded. To the right of the axis, there is a letter y.
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12.
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For the interval expressed in the number line, write it using set-builder notation and interval notation.
Graph of an interval on a number line. There is a right parenthesis at 3.5 and everything left of that is shaded. To the right of the axis, there is a letter q.
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Technology and Tables.

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13.
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Let f(x)=x2−2x−2. Use technology to make a table of values f.
x f(x)
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14.
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Let g(x)=x2+x−2. Use technology to make a table of values g.
x g(x)
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15.
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Let g(x)=−x2−4x+1. Use technology to make a table of values g.
x g(x)
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16.
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Let h(x)=−x2+3x+3. Use technology to make a table of values h.
x h(x)
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17.
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Let F(x)=3x2−7x+4. Use technology to make a table of values F.
x F(x)
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18.
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Let G(x)=−3x2+5x+4. Use technology to make a table of values G.
x G(x)
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19.
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Let H(x)=2x2+5x+26. Use technology to make a table of values H.
x H(x)
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20.
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Let H(x)=−3x2−4x+50. Use technology to make a table of values H.
x H(x)
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Technology and Graphs.

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21.
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Use technology to make a graph of f where f(x)=x2+3x−2.
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22.
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Use technology to make a graph of f where f(x)=x2−2x−1.
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23.
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Use technology to make a graph of f where f(x)=−x2+3x+2.
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24.
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Use technology to make a graph of f where f(x)=−x2+x+2.
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25.
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Use technology to make a graph of f where f(x)=3x2−6x−5.
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26.
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Use technology to make a graph of f where f(x)=−3x2−8x+3.
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27.
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Use technology to make a graph of f where f(x)=−3x2+4x+49.
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28.
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Use technology to make a graph of f where f(x)=2x2−2x+41.
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Technology and Features of Quadratic Function Graphs.

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Use technology to find features of a quadratic function and its graph.
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29.
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Let K(x)=2x2+2x−1. Use technology to find the following.
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  1. The vertex is .
  2. The y-intercept is .
  3. The x-intercept(s) is/are .
  4. The domain of K is .
  5. The range of K is .
  6. Calculate K(−1). .
  7. Solve K(x)=4.
  8. Solve K(x)>4.
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30.
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Let f(x)=x2−x−3. Use technology to find the following.
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  1. The vertex is .
  2. The y-intercept is .
  3. The x-intercept(s) is/are .
  4. The domain of f is .
  5. The range of f is .
  6. Calculate f(4). .
  7. Solve f(x)=−1.
  8. Solve f(x)<−1.
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31.
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Let g(x)=1.6x2−1.6x−0.9. Use technology to find the following.
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  1. The vertex is .
  2. The y-intercept is .
  3. The x-intercept(s) is/are .
  4. The domain of g is .
  5. The range of g is .
  6. Calculate g(4). .
  7. Solve g(x)=3.
  8. Solve g(x)≤3.
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32.
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Let g(x)=−0.5x2+3.4x−3.7. Use technology to find the following.
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  1. The vertex is .
  2. The y-intercept is .
  3. The x-intercept(s) is/are .
  4. The domain of g is .
  5. The range of g is .
  6. Calculate g(3). .
  7. Solve g(x)=−2.
  8. Solve g(x)≤−2.
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33.
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Let h(x)=x22−3.6x+3.6. Use technology to find the following.
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  1. The vertex is .
  2. The y-intercept is .
  3. The x-intercept(s) is/are .
  4. The domain of h is .
  5. The range of h is .
  6. Calculate h(3). .
  7. Solve h(x)=2.
  8. Solve h(x)≥2.
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34.
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Let F(x)=x24+1.4x+0.8. Use technology to find the following.
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  1. The vertex is .
  2. The y-intercept is .
  3. The x-intercept(s) is/are .
  4. The domain of F is .
  5. The range of F is .
  6. Calculate F(3). .
  7. Solve F(x)=8.
  8. Solve F(x)≥8.
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Applications.

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35.
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An object was launched from the top of a hill with an upward vertical velocity of 140 feet per second. The height of the object can be modeled by the function h(t)=−16t2+140t+150, where t represents the number of seconds after the launch. Assume the object landed on the ground at sea level. Find the answer using graphing technology.
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The object’s height was when it was launched.
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36.
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An object was launched from the top of a hill with an upward vertical velocity of 160 feet per second. The height of the object can be modeled by the function h(t)=−16t2+160t+100, where t represents the number of seconds after the launch. Assume the object landed on the ground at sea level. Find the answer using graphing technology.
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Use a table to list the object’s height within the first second after it was launched, at an increment of 0.1 second. Fill in the blanks. Round your answers to two decimal places when needed.
Time in Seconds Height in Feet
0.1
0.2
0.3
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37.
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An object was launched from the top of a hill with an upward vertical velocity of 180 feet per second. The height of the object can be modeled by the function h(t)=−16t2+180t+250, where t represents the number of seconds after the launch. Assume the object landed on the ground at sea level. Use technology to find the answer.
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The object was feet in the air 10 seconds after it was launched.
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38.
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An object was launched from the top of a hill with an upward vertical velocity of 190 feet per second. The height of the object can be modeled by the function h(t)=−16t2+190t+150, where t represents the number of seconds after the launch. Assume the object landed on the ground at sea level. Find the answer using technology.
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seconds after its launch, the object reached its maximum height of feet.
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39.
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An object was launched from the top of a hill with an upward vertical velocity of 50 feet per second. The height of the object can be modeled by the function h(t)=−16t2+50t+300, where t represents the number of seconds after the launch. Assume the object landed on the ground at sea level. Find the answer using technology.
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seconds after its launch, the object fell to the ground at sea level.
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40.
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An object was launched from the top of a hill with an upward vertical velocity of 70 feet per second. The height of the object can be modeled by the function h(t)=−16t2+70t+250, where t represents the number of seconds after the launch. Assume the object landed on the ground at sea level. Find the answer using technology. Round your answers to two decimal places. If there is more than one answer, use a comma to separate them.
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The object was 266 feet high at the following number of seconds after it was launched: .
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41.
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In a race, a car drove through the starting line at the speed of 5 meters per second. It was accelerating at 2 meters per second squared. Its distance from the starting position can be modeled by the function d(t)=t2+5t. Find the answer using technology.
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After seconds, the car was 234 meters away from the starting position.
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42.
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In a race, a car drove through the starting line at the speed of 10 meters per second. It was accelerating at 2.5 meters per second squared. Its distance from the starting position can be modeled by the function d(t)=1.25t2+10t. Find the answer using technology.
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After seconds, the car was 700 meters away from the starting position.
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43.
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A farmer purchased 480 meters of fencing, and will build a rectangular pen with it. To enclose the largest possible area, what should the pen’s length and width be? Model the pen’s area with a function, and then find its maximum value.
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Use a comma to separate your answers.
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To enclose the largest possible area, the pen’s length and width should be meters.
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44.
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A farmer purchased 270 meters of fencing, and will build a rectangular pen along a river. This implies the pen has only 3 fenced sides. To enclose the largest possible area, what should the pen’s length and width be? Model the pen’s area with a function, and then find its maximum value.
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To enclose the largest possible area, the pen’s length and width should be meters.
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Quadratic Functions in Vertex Form.

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45.
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Find the vertex of the graph of y=4(x−8)2−6.
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46.
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Find the vertex of the graph of y=7(x−1)2+3.
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47.
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Find the vertex of the graph of y=9(x+6)2−10.
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48.
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Find the vertex of the graph of y=−10(x−8)2−1.
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49.
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Find the vertex of the graph of y=−7(x−0.2)2+7.4.
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50.
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Find the vertex of the graph of y=−4.8(x+6.8)2−4.3.
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Exercise Group.

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51.
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A graph of a function f is given. Use the graph to write a formula for f in vertex form. You will need to identify the vertex and also one more point on the graph to find the leading coefficient a.
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f(x)=
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52.
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A graph of a function f is given. Use the graph to write a formula for f in vertex form. You will need to identify the vertex and also one more point on the graph to find the leading coefficient a.
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f(x)=
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53.
🔗
A graph of a function f is given. Use the graph to write a formula for f in vertex form. You will need to identify the vertex and also one more point on the graph to find the leading coefficient a.
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f(x)=
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54.
🔗
A graph of a function f is given. Use the graph to write a formula for f in vertex form. You will need to identify the vertex and also one more point on the graph to find the leading coefficient a.
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f(x)=
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55.
🔗
A graph of a function f is given. Use the graph to write a formula for f in vertex form. You will need to identify the vertex and also one more point on the graph to find the leading coefficient a.
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f(x)=
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56.
🔗
A graph of a function f is given. Use the graph to write a formula for f in vertex form. You will need to identify the vertex and also one more point on the graph to find the leading coefficient a.
🔗
f(x)=
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Exercise Group.

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57.
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Write the vertex form for the quadratic function f, whose vertex is (4,−6) and has leading coefficient a=−9.
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58.
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Write the vertex form for the quadratic function f, whose vertex is (−2,8) and has leading coefficient a=−7.
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59.
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Write the vertex form for the quadratic function f, whose vertex is (−9,3) and has leading coefficient a=−5.
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60.
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Write the vertex form for the quadratic function f, whose vertex is (4,−1) and has leading coefficient a=−3.
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Exercise Group.

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61.
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Let F be defined by F(x)=(x+3)2+7.
  1. What is the domain of F?
  2. What is the range of F?
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62.
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Let G be defined by G(x)=(x−9)2−4.
  1. What is the domain of G?
  2. What is the range of G?
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63.
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Let G be defined by G(x)=6.5(x−4)2+1.
  1. What is the domain of G?
  2. What is the range of G?
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64.
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Let H be defined by H(x)=2.9(x+7)2−4.
  1. What is the domain of H?
  2. What is the range of H?
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65.
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Let K be defined by K(x)=9(x−1)2+9.
  1. What is the domain of K?
  2. What is the range of K?
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66.
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Let f be defined by f(x)=2(x−9)2+4.
  1. What is the domain of f?
  2. What is the range of f?
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67.
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Let g be defined by g(x)=−4(x−15)2−23.
  1. What is the domain of g?
  2. What is the range of g?
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68.
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Let g be defined by g(x)=9(x−65)2+67.
  1. What is the domain of g?
  2. What is the range of g?
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69.
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Let h be defined by h(x)=−6(x+75)2−49.
  1. What is the domain of h?
  2. What is the range of h?
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70.
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Let F be defined by F(x)=3(x+13)2−58.
  1. What is the domain of F?
  2. What is the range of F?
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Exercise Group.

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71.
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Consider the graph of the equation y=(x−2)2+7. Compared to the graph of y=x2, the vertex has been shifted units
  • ?
  • left
  • right
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and units
  • ?
  • down
  • up
🔗
.
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72.
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Consider the graph of the equation y=(x−4)2−1. Compared to the graph of y=x2, the vertex has been shifted units
  • ?
  • left
  • right
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and units
  • ?
  • down
  • up
🔗
.
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73.
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Consider the graph of the equation y=(x−61)2−66. Compared to the graph of y=x2, the vertex has been shifted units
  • ?
  • left
  • right
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and units
  • ?
  • down
  • up
🔗
.
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74.
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Consider the graph of the equation y=(x−82.9)2+63.3. Compared to the graph of y=x2, the vertex has been shifted units
  • ?
  • left
  • right
🔗
and units
  • ?
  • down
  • up
🔗
.
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75.
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🔗
Consider the graph of the equation y=(x−38)2+79. Compared to the graph of y=x2, the vertex has been shifted units
  • ?
  • left
  • right
🔗
and units
  • ?
  • down
  • up
🔗
.
🔗
76.
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🔗
Consider the graph of the equation y=(x+38)2−57. Compared to the graph of y=x2, the vertex has been shifted units
  • ?
  • left
  • right
🔗
and units
  • ?
  • down
  • up
🔗
.
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Three Forms of Quadratic Functions.

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77.
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The quadratic expression (x−2)2−1 is written in vertex form.
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  1. Write the expression in standard form.
  2. Write the expression in factored form.
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78.
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The quadratic expression (x−3)2−25 is written in vertex form.
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  1. Write the expression in standard form.
  2. Write the expression in factored form.
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79.
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The quadratic expression (x−4)2−81 is written in vertex form.
🔗
  1. Write the expression in standard form.
  2. Write the expression in factored form.
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80.
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The quadratic expression (x−1)2−36 is written in vertex form.
🔗
  1. Write the expression in standard form.
  2. Write the expression in factored form.
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Factored Form and Intercepts.

🔗
81.
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The formula for a quadratic function g is g(x)=(x−4)(x+2).
  1. The y-intercept is .
  2. The x-intercept(s) is/are .
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82.
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The formula for a quadratic function G is G(x)=(x−6)(x+9).
  1. The y-intercept is .
  2. The x-intercept(s) is/are .
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83.
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The formula for a quadratic function f is f(x)=8(x−4)(x+9).
  1. The y-intercept is .
  2. The x-intercept(s) is/are .
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84.
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The formula for a quadratic function H is H(x)=−9(x+3)(x+1).
  1. The y-intercept is .
  2. The x-intercept(s) is/are .
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85.
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🔗
The formula for a quadratic function K is K(x)=−7x(x−9).
  1. The y-intercept is .
  2. The x-intercept(s) is/are .
🔗
86.
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🔗
The formula for a quadratic function h is h(x)=−5x(x−3).
  1. The y-intercept is .
  2. The x-intercept(s) is/are .
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87.
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🔗
The formula for a quadratic function G is G(x)=−3(x+4)(x+4).
  1. The y-intercept is .
  2. The x-intercept(s) is/are .
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88.
🔗
🔗
The formula for a quadratic function f is f(x)=−(x−9)(x−9).
  1. The y-intercept is .
  2. The x-intercept(s) is/are .
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89.
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🔗
The formula for a quadratic function g is g(x)=2(6x−1)(4x−9).
  1. The y-intercept is .
  2. The x-intercept(s) is/are .
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90.
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🔗
The formula for a quadratic function g is g(x)=4(4x−9)(7x−6).
  1. The y-intercept is .
  2. The x-intercept(s) is/are .
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