Answer
The function has a minimum value of $-17$.
Work Step by Step
The given function is
$\Rightarrow f(x)=4x^2-28x+32$
Add $17$ to each side.
$\Rightarrow f(x)+17=4x^2-28x+32+17$
Simplify.
$\Rightarrow f(x)+17=4x^2-28x+49$
Factor out $4$ from the right side.
$\Rightarrow f(x)+17=4(x^2-7x+\frac{49}{4})$
Write the right side as the square of a binomial.
$\Rightarrow f(x)+17=4(x-\frac{7}{2})^2$
Write in vertex form.
$\Rightarrow f(x)=4(x-\frac{7}{2})^2-17$
The vertex is $(\frac{7}{2},-17)$. Because $a$ is positive $(a=4)$, the parabola opens up and the $y-$coordinate of the vertex is the minimum value.
Hence, the function has a minimum value of $-17$.
