Vertex
General equation of a parable
[tex]f(x)=a(x-h)^2+k[/tex]where
[tex](h,k)[/tex]is the vertex
for our equation
[tex]f(x)=-2(x+5)^2-2[/tex]the vertex is
[tex](-5,-2)[/tex]axis of symmetry
axis of symmetry is the x value of the vertex, then
[tex]x=-5[/tex]Graph
we replace values on x to find points of the parable
for example x=0
[tex]\begin{gathered} f(0)=-2(0+5)^2-2 \\ f(0)=-2(5)^2-2 \\ f(0)=-2\times25-2 \\ f(0)=-50-2 \\ f(0)=-52 \end{gathered}[/tex]x=-10
[tex]\begin{gathered} f(-10)=-2(-10+5)^2-2 \\ f(-10)=-2(-5)^2-2 \\ f(-10)=-2\times25-2 \\ f(-10)=-50-2 \\ f(-10)=-52 \end{gathered}[/tex]our three points
[tex]\begin{gathered} (-5,-2) \\ (0,-52) \\ (-10,-52) \end{gathered}[/tex]