Question

The area of the triangle formed by points of intersection of parabola y=a(x−3)(x+2) with the coordinate axes is 10. Find a if it is known that parabola opens upward.

Answer 1

Answer: a = 4

Step-by-step explanation: Area of a triangle is calculated as: $A_{t}=\frac{b.h}{2}$.

The triangle formed by the parabola has base (b) equal to the distance between the points where the graph touches x-axis and height (h) is the point where graph touches the y-axis.

The points on the x-axis are the roots of the quadratic equation:

a(x-3)(x+2)=0

(x-3)(x+2)=0

x - 3 = 0

x = 3

or

x + 2 = 0

x = -2

So, base is the distance between (-2,0) and (3,0).

Since they are in the same coordinate, distance will be:

b = 3 - (-2)

b = 5

Area of the triangle is 10. So constant a is

$10=\frac{5.a}{2}$

5a = 10.2

a = 4

The constant a of the function y = a(x-3)(x+2) is 4.