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,

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

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step-by-step explanation:

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