Chapter 11 - Section 11.5 - Quadratic Functions and Their Graphs - Exercise Set - Page 810: 29

Please see the graph.

$f(x)=10*(x+4)^2-6$ Axis of symmetry: $f(x)=10*(x+4)^2-6$ $f(x)=10*(x- -4)^2-6$ $x=-4$ Vertex: $f(x)=10*(x+4)^2-6$ $f(-4)=10*(-4+4)^2-6$ $f(-4)=10*(0)^2-6$ $f(-4)=10*0-6$ $f(-4)=0-6$ $f(-4)=-6$ The red line is the function $f(x)=10*(x+4)^2-6$, the orange line is the axis of symmetry $x=-4$, and the dot at $(-4, -6)$ is the vertex of the function.