Question

Which of the following quadratic function has a maximum value of 16?

Answer 1

When a quadratic is in the form $y=ax^2+bx+c$, your min/max = $c-\frac{b^2}{4a}$.

When a quadratic is in the form $y=a(x-h)^2+k$, your min/max = $k$.

Let's take a look at all of these answers.

1) y = x² + 16
#1 in is general form. (y=ax²+bx+c)
a=1, b=0, c=16.
$16-\frac{0^2}{4*1}=\boxed{16}$
So we know that the min/max is 16. We don't know which it is, though.

2) y = -x² + 16
I've rearranged this equation to general form.
As you can see, this will have the same outcome as the previous, with the min/max being 16.

3) y = (x-16)²
This is in y = (x-h)² + k form, but there is no k, thus the min/max k = 0.
This is also true for 4) y = (x+16)².

Let's go back to #1 and #2.
While vertex form y = (x-h)² + k makes finding the vertex easy, it is a lot easier to know whether we have a minimum or a maximum in y = ax² + by + c form.

Simply put: If a is positive, the parabola opens upwards and we have a min.
If a is negative, it opens downwards and we have a max.

We want a maximum value of 16, so that means negative a.
Thus our answer is 2) y = -x² + 16