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If a standard parabola (y = x^2) is moved such that the new vertex is (-2,-4) and the...

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kristenmarieb... | Student, Grade 10 | Valedictorian

Posted September 22, 2013 at 1:38 AM via web

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If a standard parabola (y = x^2) is moved such that the new vertex is (-2,-4) and the point (1,1) becomes (1,-5), what will the equation of the new parabola be?

a)  y = -(3(x+2))^2 - 4

b)  y = -(3(x - 2))^2 + 4

c)  y = -1/3(x + 2)^2 - 4

d)  y = -1/3(x - 2)^2 + 4

e)  y = -(x+2/3)^2 - 4

f)  y = -(x-2/3)^2 - 4

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted September 22, 2013 at 2:38 AM (Answer #1)

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We are given a parabola with vertex (-2,-4) that includes the point (1,-5), and we are asked to find the equation for the parabola.

The vertex form for a parabola is `y=a(x-h)^2+k` where the vertex is at (h,k) and a indicates whether the parabola opens up or down and how wide. (From the parent function `y=x^2` h performs a horizontal translation, k a vertical translation, and a a dilation.)

Here h=-2, k=-4 so it remains to find a:

Substitute the known values from the point (1,-5),h, and k into the equation and solve for a:

`-5=a(1+2)^2-4`

`-1=9a`

`a=-1/9`

So the equation is `y=-1/9(x+2)^2-4` .

Scanning the list for an equivalent equation we note that b,d,and f cannot be the answer as h is wrong.

Answer A has a=-9, answer C has `a=-1/3` so these are wrong leaving only (e).

Rewrite `-1/9(x+2)^2-4` as `-(1/9(x+2)^2)-4=-(((x+2)^2)/9)-4`

`=-((x+2)/3)^2-4` which is answer (e).

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