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Hi there :) So I am studying for my Pre Calculus 11 Final Exam, and I am stuck on...

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mclark2000 | eNotes Newbie

Posted June 6, 2013 at 3:44 PM via web

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Hi there :)

So I am studying for my Pre Calculus 11 Final Exam, and I am stuck on questions like these:

 

Rewrite y = x^2 - 3x + 5 in the form y = a(x - p)^2 + q

 

Please show the steps! :D 

Thanks for your help in advance :)

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted June 6, 2013 at 3:55 PM (Answer #1)

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You need to complete the square `x^2 - 3x` using the following formula, such that:

`a^2 - 2ab + b^2 = (a - b)^2 `

Considering `a^2 = x^2` and `3x = 2ab` yields:

`{(a^2 = x^2),(3x = 2ab):} => {(a = x),(3x = 2xb):} => {(a = x),(b = 3/2):} `

Since the missing term is `b^2` , you need to complete the square by adding `(3/2)^2` , but to preserve the equation, you also need to subtract `(3/2)^2` , such that:

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

`y = (x - 3/2)^2 + 5 - 9/4`

`y = (x - 3/2)^2 + 5 - 9/4 => y = (x - 3/2)^2 + 11/4`

Hence, converting the general form of quadratic `y = x^2 - 3x + 5` into vertex form `y = a(x - p)^2 + q` , yields` y = (x - 3/2)^2 + 11/4` , where `a = 1, p = 3/2, q = 11/4.`

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