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What is the equation of a parabola with a vertex of (5,-2) and point of (6,1)? Include...

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kristenmarieb... | Student, Grade 10 | (Level 1) Valedictorian

Posted June 21, 2013 at 2:23 AM via web

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What is the equation of a parabola with a vertex of (5,-2) and point of (6,1)?

Include "y=" as part of your answer. 

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llltkl | College Teacher | (Level 3) Valedictorian

Posted June 21, 2013 at 3:23 AM (Answer #1)

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The equation of a parabola in vertex form is `y=a(x-h)^2+k` where (h,k) is the vertex.

Here, vertex is (5,-2)

We also have a point (x,y) = (6,1). So let's plug these in and find 'a'.

`1=a(6-5)^2+(-2)`

`rArr 1=a(1)^2-2`

`rArr 1=a-2`

`rArr a=3`

Here a is positive so the parabola opens upwards like a regular "U".

Now, rewrite the equation `y=a(x-h)^2+k`

Substituting a=3, h=5 and k=-2 we get:

`y=3(x-5)^2+(-2)`

`rArr y=3(x^2-10x+25)-2`

`rArr y=3x^2-30x+75-2`

`rArr y=3x^2-30x+73`

Therefore, the equation of the parabola is `y=3x^2-30x+73`

 

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