Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
Chapter Closure

Exercise 139 Page 314

You will need to substitute 0 and solve for a variable twice before graphing.

y-intercept: (0,- 15)
x-intercepts: (- 3,0) and (5,0)

To determine the x- and y-intercepts of a line, we need to substitute 0 for one variable, solve, then repeat for the other variable.

Finding the x-intercepts

To find where a function intercepts the x-axis, the function can be set equal to zero. Then, the x-values that satisfy the equation are the zeros of the function, also called the roots. Thus, to find the x-intercepts of the given function we can set it equal to zero and solve for x. x^2-2x-15=0 To solve the above equation for x, we can start by factoring. Then we will use the Zero Product Property.

Factoring

To factor a trinomial with a leading coefficient of 1, think of the process as multiplying two binomials in reverse. Let's start by taking a look at the constant term.

x^2-2x- 15=0 In this case we have - 15. This is a negative number, so for the product of the constant terms in the factors to be negative, these constants must have the opposite sign — one positive and one negative.

Factor Constants Product of Constants
1 and - 15 - 15
- 1 and 15 - 15
-3 and 5 - 15
3 and - 5 - 15

Next, let's consider the coefficient of the linear term. x^2-2x-15=0 For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, - 2.

Factors Sum of Factors
1 and - 15 - 14
-1 and 15 14
3 and - 5 - 2
- 3 and 5 2

We found the factors whose product is - 15 and whose sum is - 2. x^2- 2x-15=0 ⇕ (x+3)(x-5)=0 Now, the equation is written in a factored form.

Zero Product Property

Since the equation is already written in factored form, we can now use the Zero Product Property.
(x+3)(x-5)=0
lcx+3=0 & (I) x-5=0 & (II)
lx=- 3 x-5=0
lx_1=- 3 x_2=5
We found that x=- 3 or x=5. Thus, the coordinates of the x-intercepts are (- 3, 0) and (5,0).

Finding the y-intercept

To find the y-intercept, consider the point where the graph of the equation crosses the y-axis. The x-value of the ( x, y) coordinate pair at the y-intercept is 0. Therefore, substituting 0 for x will give us the y-intercept.
y=x^2-2x-15
y=( 0)^2-2( 0)-15
y=(0)-2(0)-15
y=- 15
A y-intercept of - 15 means that the graph passes through the y-axis at the point (0, - 15).