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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.

Sagot :

MrRoyal

Answer:

[tex]y = 4(x - 4)^2 - 2[/tex]

or

[tex]y=4x^2 -32x + 62[/tex]

Step-by-step explanation:

Given

[tex]V = (4,-2)[/tex] --- vertex

[tex]P = (2,14)[/tex] --- point

Required

The equation of the parabola

The equation of the parabola is of the form

[tex]y = a(x - h)^2 + k[/tex]

Where

[tex]V (4,-2) = (h,k)[/tex] ---- the vertex

So, we have:

[tex]y = a(x - h)^2 + k[/tex]

[tex]y = a(x - 4)^2 - 2[/tex]

In [tex]P = (2,14)[/tex], we have:

[tex](x,y) = (2,14)[/tex]

Substitute [tex](x,y) = (2,14)[/tex] in [tex]y = a(x - 4)^2 - 2[/tex]

[tex]14 = a(2 - 4)^2 - 2[/tex]

[tex]14 = a(- 2)^2 - 2[/tex]

[tex]14 = a*4 - 2[/tex]

[tex]14 = 4a - 2[/tex]

Collect like terms

[tex]4a = 14 +2[/tex]

[tex]4a = 16[/tex]

Divide both sides by 4

[tex]a= 4[/tex]

So:

[tex]y = a(x - 4)^2 - 2[/tex] becomes

[tex]y = 4(x - 4)^2 - 2[/tex]

Open bracket to express the equation in standard form

[tex]y=4(x^2 -8x + 16) - 2[/tex]

[tex]y=4x^2 -32x + 64 - 2[/tex]

[tex]y=4x^2 -32x + 62[/tex]

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