Given:
[tex]y=h(x)\text{ where }h(x)=f(-x)\text{ and }f(x)=x^2-4x+12.[/tex]Required:
We need to find the turning point of y=h(x).
Explanation:
Replace x =-x in the function f(x) to find h(x).
[tex]f(-x)=(-x)^2-4(-x)+12.[/tex][tex]f(-x)=x^2+4x+12.[/tex]Replace f(-x)=y in the equation.
[tex]y=x^2+4x+12.[/tex][tex]y=x^2+2\times2x+4+8.[/tex][tex]y=x^2+2\times2x+2^2+8.[/tex][tex]Use\text{ }(a^2+2ab+b^2)=(a+b)^2.[/tex][tex]y=(x+2)^2+8.[/tex]Which is of the form
[tex]y=a(x-h)^2+k.[/tex]where a=1, h=-2, and k=8.
The point (h,k)=(-2,8) is the turning point.
Final answer:
The turning point of y=h(x) is (-2,8).
