Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
4. Section 9.4
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Exercise 134 Page 531

Practice makes perfect
a Let's make the x→ y table.
x x^2-2x-3 y
- 1 ( - 1)^2-2( - 1)-3 0
0 0^2-2( 0)-3 - 3
1 1^2-2( 1)-3 - 4
2 2^2-2( 2)-3 - 3
3 3^2-2( 3)-3 0

Now we can draw the graph.

b Examining the graph, we see that it is a parabola which means the graph corresponds to a second degree equation. We also see that it has two x-intercepts, x=2 and x=5. With this information, we can write the equation in factored form.
y=a(x-2)(x-5)To find a, we need one more point through which the graph passes. Examining the diagram, we see that it's vertex is at (3.5,- 2). By substituting this point into the equation, we can solve for a.
y=a(x-2)(x-5)
- 2=a( 3.5-2)( 3.5-5)
Solve for x
- 2=a(1.5)(- 1.5)
- 2=- 2.25a
2=2.25a
2.25a=2
a=2/2.25
a=8/9
A possible equation is y= 89(x-2)(x-5).
c If a second degree equation can be factored, we are able to rewrite the coefficient of the x-term as a sum of two numbers, a and b, whose product equals the expression's constant.
y=x^2+(a+b)x+ab ⇕ y=(x+a)(x+b) From the equation, we can identify what this sum and product must be. cccccc y&=&x^2 & + & 8 & -7pt x & + & 7 y&=&x^2 & + & (a+b) & -7pt x & + & abLet's list all of the ways that 7 can be factored and investigate which pair of numbers, a and b, that sum to 8. Notice that because both the constant and the product are positive, both a and b must be positive. c|c|c integer & ab & a+b [0.3em] 7 & 1(7) & 8 When a and b are 1 and 7, we have factored the equation. y=x^2+8x+7 ⇔ y=(x+1)(x+7) Having factored the equation, we can find the x-intercepts by setting y equal to 0 and solving for x.
y=(x+1)(x+7)
0=(x+1)(x+7)
Solve for x
lcx+1=0 & (I) x+7=0 & (II)
lx=- 1 x+7=0
lx_1=- 1 x_2=-7
The function has two x-intercepts at x=- 1 and x=- 7. Since this is a second degree equation, we can find the vertex by calculating the average of these intercepts. average: - 1+(- 7)/2=- 4 When we know the vertex x-value we can find its y-value by substituting x=4 into the function.
y=x^2+8x+7
y=( - 4)^2+8( - 4)+7
Simplify right-hand side
y=16+8(- 4)+7
y=16-32+7
y=- 9
The vertex is at (- 4,- 9). With this information, we can draw the graph.
d To find the maximum height and how long it takes to reach this height, we have to find the vertex. Notice that the function is a second degree equation. This means we can find the vertex if we find two points with the same y-coordinate such as its x-intercepts.
y=128x-16x^2
0=128x-16x^2
Solve for x
0=16x(8-x)
0=x(8-x)
lcx=0 & (I) 8-x=0 & (II)
lx=0 - x=- 8
lx_1=0 x_2=8
The function intercepts the x-axis at x=0 and x=8. By calculating the average of these, we can find the x-value of the vertex. average: 0+8/2=4 The rocket reaches its maximum height after 4 seconds. By substituting this value into the function, we can determine the rocket's maximum height.
y=128x-16x^2
y=128( 4)-16( 4)^2
Simplify right-hand side
y=128(4)-16(16)
y=512-256
y=256
The maximum height is 256 feet.