To answer the question about the vertex and [tex]$x$[/tex]-intercepts of the graph of the quadratic equation [tex]\( y = (x+4)(x-2) \)[/tex], let's proceed step-by-step.
### Finding the [tex]$x$[/tex]-intercepts
1. The [tex]$x$[/tex]-intercepts of the graph are the points where [tex]\( y \)[/tex] is equal to 0.
2. Set the equation equal to zero: [tex]\( (x+4)(x-2) = 0 \)[/tex].
3. Solve for [tex]\( x \)[/tex] by setting each factor to zero:
- [tex]\( x + 4 = 0 \)[/tex] ⟹ [tex]\( x = -4 \)[/tex]
- [tex]\( x - 2 = 0 \)[/tex] ⟹ [tex]\( x = 2 \)[/tex]
4. Therefore, the [tex]$x$[/tex]-intercepts are [tex]\( (-4, 0) \)[/tex] and [tex]\( (2, 0) \)[/tex].
### Finding the Vertex
1. To convert the given quadratic equation into standard form, expand it:
[tex]\[
y = (x+4)(x-2) = x^2 - 2x + 4x - 8 = x^2 + 2x - 8
\][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -8 \)[/tex].
2. The formula for the [tex]\( x \)[/tex]-coordinate of the vertex of a parabola given by [tex]\( y = ax^2 + bx + c \)[/tex] is [tex]\( x = -\frac{b}{2a} \)[/tex]:
[tex]\[
x = -\frac{2}{2 \cdot 1} = -1
\][/tex]
3. Substitute [tex]\( x = -1 \)[/tex] back into the quadratic equation to find the [tex]\( y \)[/tex]-coordinate:
[tex]\[
y = 1(-1)^2 + 2(-1) - 8 = 1 \cdot 1 - 2 - 8 = 1 - 2 - 8 = -9
\][/tex]
4. Thus, the vertex is [tex]\( (-1, -9) \)[/tex].
### Final Answers
- The [tex]$x$[/tex]-intercepts are [tex]\(( -4, 0)\)[/tex] and [tex]\(( 2, 0)\)[/tex].
- The vertex is [tex]\((-1, -9)\)[/tex].
So, the selected answers from the given options are:
- C. [tex]$x$[/tex]-intercepts: [tex]$(-4,0),(2,0)$[/tex]
- B. Vertex: [tex]$(-1,-9)$[/tex]
