Question

The graph of a function is a parabola that has a minimum at the point (-3,9). Which equation could represent the function?

A. g(x)=3(x-3)^2+9

B. g(x)=-1/2(x-3)^2+9

C. g(x)=-(x+3)^2+9

D. g(x)=2(x+3)^2+9

Answer 1

The answer is D

Explanation:

I just took the test and this was the correct answer, but the person who answered first deserves the credit because he helped me out.

Answer 2

`g(x)=2(x+3)^2+9`, which corresponds to answer D.

Explanation:

Recall that the maximum or minimum of a parabola is always located at its vertex. Notice that all quadratic functions listed are given in what is called "vertex" form, since they explicitly show the vertex coordinates in their formulation:

`f(x)=a(x-x_v)^2+y_v` where
`x_v`, and
`y_v` stand for the x and y coordinates respectively of the vertex.

Examining all four options, we notice that in all four cases listed, the
`y_v` is correct (equal to positive 9), so we proceed to examine what the
`(x-x_v)^2` expression should look like for
`x_v=-3`:

Notice that when replace
`x_v` with "-3", we end up with a sign change:

`(x-x_v)^2\\(x-(-3))^2=(x+3)^2`

Therefore we find that the last two options can be candidates. Then we recall that the question also states that this should be a minimum. Then, for a parabola to have a minimum, its branches must go upwards, and this corresponds to a case in which the factor
`a` (leading coefficient) multiplying (x-x_v)^2 is positive. So in looking for such condition, we find that the very last option is the only one that verifies such.

Then,
`g(x)=2(x+3)^2+9` is the option we select.