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Changing a Quadratic Function Into a Vertex Form Correct?Consider the following...

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maythany | eNoter

Posted December 18, 2012 at 3:55 AM via web

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Changing a Quadratic Function Into a Vertex Form Correct?

  1. Consider the following quadratic function: y = −3x2 + 2x + 1
  1. Write the quadratic function in the vertex form

Here's what I did:

y=-3x^2+2x+1

=3(x^2+2/3x)+1

=3(x^2+2/3x+1/9-1/9)+1

=3(x^2+2/3x+1/9)+1-(1/3)

=3(x^2+2/3x+1/9)-2/3

=3(x+1/3)^2-(2/3)

 

I'm wondering if I did this correctly?

Thanks,

 

MB

 

1 Answer | Add Yours

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted December 18, 2012 at 6:23 AM (Answer #1)

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You need to convert the given standard form of quadratic equation into vertex form, such that:

`y = -3x^2 + 2x + 1`

`y = -3(x^2 - 2/3*x) + 1`  (factor out -3 between the first and second terms)

You need to complete the square `x^2 - 2/3*x`  using the following formula `(a- b)^2 = a^2 - 2ab + b^2` , such that:

`a^2 = x^2 => a = x`

`-2ab = -2*(1/3)*x => -2x*b = -2*x*(1/3) => b = -1/3`

`b^2 = 1/9`

Hence, you need to complete the square adding `1/9`  such that:

`y = -3(x^2 - 2/3*x + 1/9) + 1`

Adding `1/9,`  the equation changes, hence, you need to subtract 1/9 to prevent the changing of original equation, such that:

`y = -3(x^2 - 2/3*x + 1/9 - 1/9) + 1`

Converting the expansion `x^2 - 2/3*x + 1/9`  into binomial `(x^2 - 1/3)^2`  yields:

`y = -3((x^2 - 1/3)^2 - 1/9) + 1`

`y = -3(x^2 - 1/3)^2 + 3/9 + 1 => y = -3(x^2 - 1/3)^2 + 1/3 + 1`

You need to bring the terms `1/3 + 1`  to a common denominator, such that:

`1/3 + 1 = (1 + 3)/3 = 4/3 y = -3(x^2 - 1/3)^2 + 4/3`

Hence, converting the standard form of quadratic equation `y = -3x^2 + 2x + 1`  into the vertex form yields `y = -3(x^2 - 1/3)^2 + 4/3.`

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