Question

question 7 of 10
what are the vertex and x-intercepts of the graph of the function below?
y=(x + 4)(x - 2)

a. vertex: (1, 9); intercepts: x = -4, -2
b. vertex: (1, -9); intercepts: x = 4, -2
c. vertex: (-1, -9); intercepts: x = -4, 2
d. vertex: (-4, 2); intercepts: x = 3, -3

Explanation:

Step1: Find x-intercepts

To find the x-intercepts, set \( y = 0 \):
\( 0=(x + 4)(x - 2) \)
Using the zero - product property, if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \). So \( x+4 = 0 \) or \( x - 2=0 \).
Solving \( x + 4=0 \) gives \( x=-4 \), and solving \( x - 2 = 0 \) gives \( x = 2 \). So the x-intercepts are \( x=-4 \) and \( x = 2 \).

Step2: Find the vertex

First, expand the function \( y=(x + 4)(x - 2) \):
\( y=x^{2}-2x + 4x-8=x^{2}+2x - 8 \)
For a quadratic function in the form \( y=ax^{2}+bx + c \) (here \( a = 1 \), \( b = 2 \), \( c=-8 \)), the x-coordinate of the vertex is given by \( x=-\frac{b}{2a} \).
Substitute \( a = 1 \) and \( b = 2 \) into the formula: \( x=-\frac{2}{2\times1}=-1 \)
Now find the y-coordinate of the vertex by substituting \( x=-1 \) into the function \( y=x^{2}+2x - 8 \):
\( y=(-1)^{2}+2\times(-1)-8=1 - 2-8=-9 \)
So the vertex is \( (-1,-9) \).

Answer:

C. Vertex: \((-1, -9)\); Intercepts: \(x = -4, 2\)