Mathematics: Find the equation of the parabola which has the given vertex V, which passes through the given point P, and which has the specified axis of symmetry. V(4,−2),P(2,14), vertical axis of symmetry.

Find the equation of the parabola which has the

MathChic68

03.07.2021

y = 4(x - 4)^2 - 2

y=4x^2 -32x + 62

Step-by-step explanation:

Given

V = (4,-2) --- vertex

P = (2,14) --- point

Required

The equation of the parabola

The equation of the parabola is of the form

y = a(x - h)^2 + k

Where

V (4,-2) = (h,k) ---- the vertex

So, we have:

y = a(x - h)^2 + k

y = a(x - 4)^2 - 2

In P = (2,14) , we have:

(x,y) = (2,14)

Substitute (x,y) = (2,14) in y = a(x - 4)^2 - 2

14 = a(2 - 4)^2 - 2

14 = a(- 2)^2 - 2

14 = a*4 - 2

14 = 4a - 2

Collect like terms

4a = 14 +2

4a = 16

Divide both sides by 4

a= 4

So:

y = a(x - 4)^2 - 2 becomes

y = 4(x - 4)^2 - 2

Open bracket to express the equation in standard form

y=4(x^2 -8x + 16) - 2

y=4x^2 -32x + 64 - 2

y=4x^2 -32x + 62

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