## Precalculus: Mathematics for Calculus, 7th Edition

The graph of $f(x)=3(x-2)^{2}-6$ is a parabola that opens __upward__, with its vertex at (_2_ , _-6_), and $f(2)=$ __-6__ is the __minimum__ value of $f$.
With all of this, use the standard form $f(x)=a(x-h)^{y}+k$. a) The parabola opens upward if $\gt$ 0 or downward if $\lt$ 0. The a in this function is 3, which is $\gt$ 0 and opens upward. b) The vertex is the (h,k) of a function. In this function, it is (2,-6). c)You substitute the 2 in the function for x and solve: $f(x)=3(x-2)^{2}-6$ $f(2)=3(2-2)^{2}-6$ $f(2)=3(0)^{2}-6$ $f(2)=3(0)-6$ $f(2)=0-6$ $f(x)=-6$ d)The maximum or minimum value of $f$ occurs at $x=h$. If $\gt$ 0, then the minimum value of $f$ is $f(h)=k$. If $\lt$ 0, then the maximum value of $f$ is $f(h)=k$.