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

The area of the triangle formed by points of intersection of parabola y=a(x-3)(x+2) with the coordinate axes is 10. The constant (a) of the function is 4.

How to find the if the parabola is open?

The triangle formed by the parabola has a base equal to the distance between the points where the graph touches the x-axis and height (h) is the point where the 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, the base is the distance between (-2,0) and (3,0).

Since they are in the same coordinate, the distance here;

b = 3 - (-2)

b = 5

The area of the triangle is 10. So

5a = 10 x 2

a = 4

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

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