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Section A.4 Chapter 4 Applications of Quadratic Models

Subsection A.4.1 Quadratic Formula

Subsubsection A.4.1.1 Simplify square roots

Be careful when simplifying radicals.
Example A.4.1.
Can you simplify the first expression into the second expression? (Decide whether the expressions are equivalent.)
  1. Is   4+x2   equivalent to   2+x?
  2. Is   x29   equivalent to   x3   for  x0?
  3. Is   w3   equivalent to   w3?
Solution.
  1. If the expressions are equivalent, they must be equal for every value of the variable. Let’s test with x=3. Then
    4+x2=4+9=133.6but         2+x=2+3=5
    No, the expressions are not equivalent.
  2. Because (x3)2=x3x3=x232=x29, it is also true that x29=x3. Yes, the expressions are equivalent.
  3. Let w=16. Then
    w3=163=133.6but   w3=16341.7=2.3
    No, the expressions are not equivalent.
Decide whether the expressions are equivalent. Assume all variables are positive.
Checkpoint A.4.2.
b281  and  b9
Answer.
No
Checkpoint A.4.3.
64x2y2  and  8xy
Answer.
Yes
Checkpoint A.4.4.
64+x2y2  and  8+xy
Answer.
No
Checkpoint A.4.5.
c2+d24b2  and  c2+d22b
Answer.
Yes

Subsubsection A.4.1.2 Use the order of operations

In the order of operations, simplifying radicals comes after what’s inside parentheses (or fraction bars) and before products and quotients.
Example A.4.6.
Simplify each expression. Do not use a calculator!
  1. 4642
  2. 2(3163(27))
  3. 63277(5)3
Solution.
  1. We start by simplifying the numerator.
    4642=482Evaluate the radical, then subtract.=42=2Reduce the fraction.
  2. We start by simplifyng what’s inside parentheses.
    2(3163(27))Evaluate the radicals.=2(3481)Simplify inside the parentheses.=2(129)=6
  3. We start by simplifying the radicand.
    63277(5)3=6327353Subtract under the radical.=6383Evaluate the radical.=63(2)=12
Example A.4.7.
Simplify each expression. Round your answer to hundredths.
  1. 8224
  2. 2+6253
Solution.
  1. Do not start with "82"! Evaluate 2 first, then multiply by 2, and subtract the result from 8. Once the numerator is simplified, divide by 4.
    On a calculator, enter
    ( 8 - 2  ) ) ÷ 4 ENTER
    and round to two decimal places:  1.29
  2. Evaluate the cube root, multiply by 6, then add the result to 2.
    On a calculator, enter
    2 + 6 MATH 4 - 25 ENTER
    and round to two decimal places:  15.54
Checkpoint A.4.8.
Simplify each expression. Do not use a calculator!
  1. 366+36
  2. 10+2(3169)
  3. 3+72936273
Answer.
  1. 3
  2. 10
  3. 23
Checkpoint A.4.9.
Simplify each expression. Round your answer to hundredths.
  1. 6+933
  2. 131203
Answer.
  1. 7.20
  2. 15.80

Subsubsection A.4.1.3 Use the quadratic formula

Example A.4.10.
  1. Use the quadratic formula to solve the equation  x=23x26
  2. Give decimal approximations to two decimal places for the solutions.
Solution.
  1. To clear fractions, multiply each term by the LCD, 6, to get
    6(x)=6(x)(23x26)6x=4x2
    Next, write the equation in standard form and identify the coefficients.
    x2+6x4=0a=1,b=6,c=4
    Substitute the coefficients into the quadratic formula and sinplify.
    x=6±624(1)(4)2(1)=6±522
    The solutions are 6+522 and 6522
  2. We can use a calculator to find decimal approximations for the solutions.
    6+5226+7.212=0.605652267.212=6.605
    To two decimal places, the solutions are 0.61 and 6.61.
Checkpoint A.4.11.
Use the quadratic formula to solve the equation x2+2x=5
Answer.
x=2±242

Subsection A.4.2 The Vertex

Subsubsection A.4.2.1 Parabolas in vertex form

We can sketch the graph of a parabola with the vertex, the y-intercept, and its symmetric point.
Example A.4.12.
Graph the equation  y=2(x15)272
Solution.
The vertex is (15,72). We find the y-intercept by setting x=0:
y=2(015)272=378
The y-intercept is (0,378). We find the x-intercepts by setting y=0:
2(x15)272=0(x15)2=36x=±6+15
The x-intercepts are (9,0) and (21,0). We plot these three points and sketch the parabola through them.
parabola
Checkpoint A.4.13.
Find the vertex, the y-intercept, and the x-intercepts of  y=2(x+1)2+3, and sketch its graph.
Answer.
parabola

Subsubsection A.4.2.2 Product of binomials

With practice, you will be able to multiply binomials mentally.
Example A.4.14.
Multiply  (x6)(x+3).
Solution.
We apply the distributive law to multiply each term of the first factor by each term of the second factor. (The "FOIL" method.)
(x6)(x+3)=xx+3x6x63=x2+3x6x18Combine like terms.=x23x18
Example A.4.15.
Expand  (2t5)2.
Solution.
Multiply  (2t5)(2t5)
(2t5)(2t5)=4t210t10t+25Combine like terms.=4t220t+25
Checkpoint A.4.16.
Multiply  (4a7)(3a+8).
Answer.
12a2+11a56
Checkpoint A.4.17.
Expand  (34b)2.
Answer.
16b224b+9

Subsubsection A.4.2.3 Solve for a parameter

We can find the equation for a parabola if we know, for example, the vertex and one other point.
Example A.4.18.
The point (6,2) lies on the graph of  y=a(x4)2+1.  Solve for a.
Solution.
Substitute 6 for x and 2 for y, then solve for a.
2=a(64)2+12=a(4)+11=4a
The solution is a=14.
Example A.4.19.
The point (2,11) lies on the graph of  y=x2+bx3.  Solve for b.
Solution.
Substitute -2 for x and 11 for y, then solve for b.
(2)2+b(2)3=1142b3=112b=10
The solution is b=5.
Checkpoint A.4.20.
The point (6,10) lies on the graph of  y=a(x+3)22.  Solve for a.
Answer.
a=43
Checkpoint A.4.21.
The point (3,8) lies on the graph of  y=x2+bx+5.  Solve for b.
Answer.
b=4
Checkpoint A.4.22.
The point (8,12) lies on the graph of  y=ax24x+36.  Solve for a.
Answer.
a=13
Checkpoint A.4.23.
The point (60,480) lies on the graph of  y=23(xh)2+120.  Solve for h.
Answer.
h=30

Subsection A.4.3 Curve Fitting

Subsubsection A.4.3.1 Points on a graph

If a curve passes through a given point, the coordinates of the point satisfy the equation of the curve.
Example A.4.24.
Write an equation to say that (3,8) lies on the graph of y=ax2+bx+c .
Solution.
Substitute 3 for x and 8 for y.
8=a(3)2+b(3)+cSimplify.8=9a3b+c
Checkpoint A.4.25.
Write an equation to say that (4,18) lies on the graph of y=ax2+bx+c .
Answer.
16a4b+c=18
Checkpoint A.4.26.
Write an equation to say that (8,0) lies on the graph of y=ax2+bx+c .
Answer.
64a+8b+c=0
Checkpoint A.4.27.
Write an equation to say that (0,5) lies on the graph of y=ax2+bx+c .
Answer.
c=5
Checkpoint A.4.28.
Write an equation to say that (60,400) lies on the graph of y=ax2+bx+c .
Answer.
3600a60b+c=400

Subsubsection A.4.3.2 Solve a 2x2 linear system

For fitting a parabola through given points, we’ll solve systems using the method of elimination.
Example A.4.29.
Solve the system by elimination.
5x2y=222x5y=13
Solution.
To eliminate the x-terms,look for the smallest integer that both 2 and 5 divide into evenly, namely, 10. Multiply the first equation by 2 and the second equation by 5.
2(5x2y=22)10x4y=445(2x5y=13)10x+25y=65
Add these new equations to obtain an equation in y.
10x 4y=4410x+25y=6510x+25y21y=21
Solve for y to find y=1. Finally, substitute y=1 into the first equation and solve for x.
5x2(1)=225x+2=22x=4
The solution to the system is (4,1).
Checkpoint A.4.30.
Solve the system by elimination.
2x9y=34x5y=7
Answer.
(3,1)
Checkpoint A.4.31.
Solve the system by elimination.
5x+2y=54x+3y=3
Answer.
(3,5)

Subsubsection A.4.3.3 Special 3x3 linear system

In this special case of solving a 3x3 system, we can eliminate c to create a 2x2 system.
Example A.4.32.
Solve the system by elimination.
a+b+c=3(1)4a2b+c=18(2)9a+3b+c=13(3)
Solution.
Eliminate c by subtracting (1) from (2), then eliminate c again by subtracting (1) from (3), to get a 2x2 system:
3a3b=158a+2b=10
Divide the first equation by 3 and the second equation by 2, then add.
ab=54a+b=54a+b5a=10
We see that a=2. Substituting a=2 into the equation ab=5, we find that b=3. Finally, we substitue a=2 and b=3 into equation (1) to find
23+c=3c=4
The solution is a=2, b=3, and c=4.
Checkpoint A.4.33.
Solve the system by elimination.
a+b+c=54a2b+c=716a+4b+c=37
Answer.
a=3, b=1, c=7

Subsection A.4.4 Quadratic Inequalities

Subsubsection A.4.4.1 Solve a linear inequality

First, let’s review solving linear inqualities.
Example A.4.34.
Solve  3x+1>7  and graph the solutions on a number line.
Solution.
3x+1>7Subtract 1 from both sides.3x>6Divide both sides by -3.x<2Reverse the direction of the inequality.
The graph of the solutions is shown below.
number line
Example A.4.35.
Solve  3<2x56  and graph the solutions on a number line.
Solution.
3<2x56Add 5 on all three sides.2<2x11Divide each side by 2.1<x112Do not reverse the inequality.
The graph of the solutions is shown below.
number line
Checkpoint A.4.36.
Solve the inequality  84x>2 
Answer.
x<52
Checkpoint A.4.37.
Solve the inequality  64x3<2 
Answer.
22x>2
Checkpoint A.4.38.
Solve the inequality  3x5<6x+7 
Answer.
x<43
Checkpoint A.4.39.
Solve the inequality  6>45b>21 
Answer.
2<b<5

Subsubsection A.4.4.2 x-intercepts of a parabola

To solve a quadratic inequality, we first find the x-intercepts of the graph. Remember that there are four different methods for solving a quadratic equation.
Example A.4.40.
Find the x-intercepts of the parabola  y=4x212
Solution.
Set y=0 and solve for x. Use extraction of roots.
4x212=04x2=12x2=3x=±3
The x-intercepts are (3,0) and (3,0), or about (1.7,0) and (1.7,0).
Example A.4.41.
Find the x-intercepts of the parabola  y=4x212x
Solution.
Set y=0 and solve for x. Factor the right side.
0=4x212x0=4x(x+3)Set each factor equal to 0.4x=0      x+3=0x=0      x=3
The x-intercepts are (0,0) and (3,0).
Example A.4.42.
Find the x-intercepts of the parabola  y=4x212x+8
Solution.
Set y=0 and solve for x. Factor the right side.
0=4x212x+80=4(x23x+2)0=4(x2)(x1)Set each factor equal to 0.x2=0      x1=0x=2      x=1
The x-intercepts are (2,0) and (1,0).
Example A.4.43.
Find the x-intercepts of the parabola  y=1212x4x2
Solution.
Set y=0 and solve for x. Use the quadratic formula.
0=4x212x+12a=4, b=12, c=12x=12±(12)24(4)(12)2(4)=12±144+968=12±2408=12±4158240=1615=415=3±152
The x-intercepts are (3+152,0) and (3152,0), or about (0.44,0) and (3.44,0).
For each Exercise, find the x-intercepts of the parabola.
Checkpoint A.4.44.
 y=2x27x+3
Answer.
(12,0), (3,0)
Checkpoint A.4.45.
 y=7x2x2
Answer.
(0,0), (72,0)
Checkpoint A.4.46.
 y=102x2
Answer.
(2.24,0), (2.24,0)
Checkpoint A.4.47.
 y=2x2+10x+3
Answer.
(0.32,0), (4.68,0)
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