A parabola has the form f(x) = a(x-2)² - 5 and passes through the point (0, 3). What is f(4)?

Use the given point to find $a$: $f(0)=a(0-2)^2-5=4a-5=3$, so $4a=8$ and $a=2$. Then $f(4)=2(4-2)^2-5=2(4)-5=3$, choice C. This agrees with symmetry: the vertex is at $x=2$, and $x=0$ and $x=4$ are each $2$ units away, so they have the same output, $3$. Choice A ($-5$) is the vertex's $y$-value, the output at $x=2$. Choice B ($-1$) results from using $a=1$ instead of $a=2$. Choice D ($11$) results from $a=4$. Both of those fail the condition $f(0)=3$.