Find the equation of a parabola with vertex on the y-axis if points P(x1, 0) and R(3, 9) belong the parabola, the area of △PQR is 40.5sq. units, and point Q has coordinates (3, 0).

1.961x^2 - 8.647.

Step-by-step explanation:

I'm assuming the area of the triangle is 4.05 unit^2.

RQ is perpendicular to the x-axis.

The length of RQ = 9 so the area of the triangle

= 1/2 *9 * PQ = 4.05

PQ = 4.05 / 4.5 = 0.9.

So the value of x1 = 3 - 0.9 = 2.2.

The vertex is on the y-axis so the equation will be of the form ax^2 + c = 0.

So, as P (2.1, 0) and  R(3, 9) are points on the parabola we have the system:

(2.1)^2a  + c = 0

(3)^2a + c = 9

4.41 a + c = 0

9a + c = 9     Subtract:

4.59a = 9

a = 1.961

and c = -4.41* 1.961 = - 8.647,

answer: i believe the answer is e

step-by-step explanation:

the formula of a slope:

we have the points (-2, 7) and (2, 3). substitute: