SAT Math: Functions That Intersect Geometric Figures

This post addresses another of the types of problems that appear near the end of an SAT Math section among the harder questions. This type of problem requires you to integrate what you know about functions with your knowledge of geometry. Most of these problems are multi-step and require you to think ahead about what you need to know in order to solve the problem and how you can figure those things out.

Let’s consider a problem:

Triangle ABC is isosceles and has a perimeter of 32.  The vertices of its base are (-6,0) and (6, 0). The parabola has its vertex at (0, -2) and it intersects the triangle at the midpoints of segments AB and AC. If the parabola has an equation in the form f(x) = ax² + k, what is the value of a?

Functions & Shapes 1

(A)  -1/3
(B)  1/3
(C)  2/3
(D)  3/4
(E)  4/3

The key to these problems normally is finding the point where the function intersects the geometric figure. If you can find that point, you can then plug it into the equation for the function and find what you are missing. In this problem, we need to figure out what the midpoint of either of the segments AB or BC is.

Let’s start by looking at the triangle. If it is isosceles, we know that the sides AB and BC (the “non-base” sides) have the same length.  The base has a length of 12, so the remaining two sides must have lengths of 10, giving you the perimeter of 32. If you drop an altitude from point A, you’ll soon realize that you have a pair of 6-8-10 right triangles. (If you don’t remember your “famous” right triangles — it helps to learn them — you can find the length of that altitude using the Pythagorean Theorem.)

Functions & Shapes 2


You can now label the coordinates of Point A and the coordinates of the midpoints of the two segments.

Functions & Shapes 3


Now you know the coordinates of two points on the graph of the parabola.  Choose one of them to plug into the function. (Note: we know that k is -2 because that is the y-coordinate of the vertex.)

f(x) = ax² + k
f(x) = ax² – 2
4 = a(3)² – 2
4 = 9a – 2
6 = 9a
6/9 = a
2/3 = a

The correct answer is Choice (C).

If you have questions about this type of problem or anything else to do with the SAT, leave a comment below or send me an email at .

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