Solution
Explanation:
Given:
(b) Equation of the axis of symmetry
[tex]\begin{gathered} x=-8 \\ x=4 \end{gathered}[/tex][tex]\begin{gathered} x=-8,x=4 \\ (x+8)(x-4)=0 \\ x^2-4x+8x-32=0 \\ x^2+4x-32=0 \\ y=x^2+4x-32 \end{gathered}[/tex]where
[tex]\begin{gathered} y=ax^2+bx+c \\ a=1,b=4,c=-32 \end{gathered}[/tex]The formula for the axis of symmetry and the x value of the vertex
[tex]x=-\frac{b^2}{2a}[/tex]Plug in the value
[tex]x=\frac{-(4)^}{2}=-2[/tex](d) To find the y value of the vertex, substitute 1 for x in the equation.
[tex]\begin{gathered} y=x^2+4x-32 \\ y=(-2)+4(-2)-32 \\ y=-2-8-32 \\ y=-42 \end{gathered}[/tex]The vertex is (-2 , -42) Since a > 0 the vertex is the minimum point and the parabola opens upward.
Hence the vertex = (-2 , -42)