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 = 2/3

Explanation:

y-intercept: y=a(2)(-3) = -6a

x-intercepts: (-2,0) , (3,0)

height = -6a

base = 5

Area = 10 = 0.5 x 5 x -6a

10 = -15 a

a = -2/3

BUT a>0 since it opens upwards, so:

a = 2/3

Answer 2

Answer: a = 4

Explanation: Area of a triangle is calculated as:
`A_(t)=(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=(5.a)/(2)`

5a = 10.2

a = 4

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