Question

A parabola with the vertex at (1.5, 12.5) inscribes a segment of length 5 on x-axis. What is the area of the triangle formed by points of intersection of parabola with the coordinate axes.

Answer 1

20 units²

Explanation:

The x-intercepts are symmetrically located around the x-coordinate of the vertex, so are at

1.5 ± 5/2 = {-1, 4}

Using one of these we can find the unknown parameter "a" in the parabola's equation (in vertex form) ...

0 = a(4 -1.5)² +12.5

0 = 6.25a +12.5 . . . . . simplify

0 = a +2 . . . . . . . . . . . divide by 6.25

-2 = a

Then the standard-form equation of the parabola is ...

y = -2(x -1.5)² +12.5 = -2(x² -3x +2.25) +12.5

y = -2x² +6x +8

This tells us the y-intercept is 8. Then the relevant triangle has a base of 5 units and a height of 8. Its area is given by the formula ...

A = (1/2)bh = (1/2)(5)(8) = 20 . . . . units²