Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
3. Section 9.3
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Exercise 118 Page 526

Practice makes perfect
a For any function, the value of its constant tells us where it intersects the y-axis. If we examine the equation, we can identify the constant.
f(x)=x^2+x - 6 This function has a constant of -6, which means it intersects the y-axis at (0,-6).
b To determine where the graph intersects the x-axis, we will factor the right-hand side. Let's use a generic rectangle and a diamond problem. We know that x^2 and -6 go into the lower-left and upper-right corners of the generic rectangle, respectively.

Now we To fill in the remaining two corners, we need two x-terms that have a sum of x and a product of -6x^2.

Notice that the product is negative. This means one factor must be positive and the other must be negative.

Product ax(bx) ax+bx Sum x?
- 6x^2 - 3x(2x) - 3x+2x - x *
- 6x^2 - 2x(3x) - 2x+3x x ✓

When one factor is -2x and the other is 3x, we get a product of -6x^2 and a sum of x. Now we can complete the diamond and generic rectangle.

To factor the right-hand side, we add each side of the generic rectangle and multiply the sums. f(x)=x^2+x-6 ⇕ f(x)=(x-2)(x+3) Now we will we set f(x) equal to 0 and then use the Zero Product Property to solve the equation and find the x-intercepts.

f(x)=(x-2)(x+3)
0=(x-2)(x+3)
(x-2)(x+3)=0
lcx-2=0 & (I) x+3=0 & (II)
lx=2 x+3=0
lx_1=2 x_2=-3

The graph intersects the x-axis at x=2 and x=- 3.

c We want to solve the given composite function if c(x)=6.
g(x)=f(x)+c(x) To determine g(x), we will substitute c(x)=6 into the given function and evaluate the right-hand side.

g(x)=f(x)+c(x)
g(x)=( x^2+x-6)+( 6)
g(x)=x^2+x-6+6
g(x)=x^2+x

d As previously noted, the y-intercept of a function is given by the constant. If a function has no constant, it means that the y-intercept is the origin. To determine the x-intercepts, we will set g(x) equal to 0, factor the right-hand side, and solve for x using the Zero Product Property.

g(x)=x^2+x
0=x^2+x
x^2+x=0
x(x+1)=0
lcx=0 & (I) x+1=0 & (II)
lx_1=0 x_2=-1

As we can see, g(x) has x-intercepts at x=0 and x=-1. Now let's compare the graphs.