Algebra 2 help... Quadratic equations

**mathhelpplease** · November 16th 2012, 05:08 PM

9th grade here! I need help with a problem.. this is it:
[Question]
Write y=a(x-h)^2+k and y=a(x-p)(x-q) in standard form. Knowing the
vertex of the graph of y=ax^2 + bx +c occurs at x=-b/2a, show that the vertex of the
graph of y=a(x-h)^2+k occurs at x=h and that the vertex of the graph of y=a(x-p)(x-q)
occurs at x=p+q/2.

This is how far I got:
y= a(x-h)^2+k
>a(x^2-2xh-h^2)+k
>y=ax^2-a2xh-ah^2+k
>y=ax^2+(-2ah)x+(-ah^2=k)

a=a b=-2ah c=ah^2+k

>Next part:
>a(x-p)(x-q)
>a(x^2-qx-px+pq)
>y=ax^2-aqx-apx+apq

I'm stuck here, how to I arrange it to be standard form?

**chiro** · November 16th 2012, 09:49 PM

Hey mathhelpplease.

It's been a while since I was in high school so can I ask what standard form is (for question 2)?

**Soroban** · November 16th 2012, 10:06 PM

Hello, mathhelpplease!

Your work seems to be correct,
. . but I'm not sure where (or why) you are stuck.
I'll walk through it for you.

(a) Write $y\,=\,a(x-h)^2+k$ and $y\,=\,a(x-p)(x-q)$ in standard form.

$y \:=\:a(x-h)^2+k$
$y\:=\:a(x^2-2hx + h^2) + k$
$y\:=\:ax^2 - 2ahx + ah^2 + k$
$y\:=\:ax^2 - 2ahx + (ah^2+k)$ .[1]

$y \:=\:a(x-p)(x-q)$
$y\:=\:a(x^2 - px - qx + pq)$
$y\:=\:ax^2 - apx - aqx + apq$
$y\:=\:ax^2 - a(p+q)x + apq$ .[2]

(b) Knowing the vertex of the graph of $y=ax^2 + bx +c$ occurs at $x=\tfrac{\text{-}b}{2a}$
show that the vertex of the graph of $y\,=\,a(x-h)^2+k$ occurs at $x=h$
and that the vertex of the graph of $y\,=\,a(x-p)(x-q)$ occurs at $x=\tfrac{p+q}{2.}$

[1] . $y\:=\:ax^2 - 2ahx + (ah^2+k) \quad\Rightarrow\quad \begin{Bmatrix}a &=& a \\ b &=& \text{-}2ah \\ c &=& ah^2+k \end{Bmatrix}$

Vertex: . $x \;=\;\frac{\text{-}b}{2a} \;=\;\frac{\text{-}(\text{-}2ah)}{2(a)} \;=\;\frac{2ah}{2a} \;=\;h$

$\text{The vertex occurs at }\,x \,=\, h.$

[2] . $y \:=\:ax^2 - a(p+q)x + apq \quad\Rightarrow\quad \begin{Bmatrix}a &=& a \\ b &=& \text{-}a(p+q) \\ c &=& apq \end{Bmatrix}$

Vertex: . $x \;=\;\frac{\text{-}b}{2a} \;=\;\frac{\text{-}[\text{-}a(p+q)]}{2(a)} \;=\;\frac{a(p+q)}{2a} \;=\;\frac{p+q}{2}$

$\text{The vertex occurs at }\,x \,=\,\tfrac{p+q}{2}$

**mathhelpplease** · November 17th 2012, 07:54 AM

Originally Posted by Soroban

Hello, mathhelpplease!

Your work seems to be correct,
. . but I'm not sure where (or why) you are stuck.
I'll walk through it for you.

$y \:=\:a(x-h)^2+k$
$y\:=\:a(x^2-2hx + h^2) + k$
$y\:=\:ax^2 - 2ahx + ah^2 + k$
$y\:=\:ax^2 - 2ahx + (ah^2+k)$ .[1]

$y \:=\:a(x-p)(x-q)$
$y\:=\:a(x^2 - px - qx + pq)$
$y\:=\:ax^2 - apx - aqx + apq$
$y\:=\:ax^2 - a(p+q)x + apq$ .[2]

[1] . $y\:=\:ax^2 - 2ahx + (ah^2+k) \quad\Rightarrow\quad \begin{Bmatrix}a &=& a \\ b &=& \text{-}2ah \\ c &=& ah^2+k \end{Bmatrix}$

Vertex: . $x \;=\;\frac{\text{-}b}{2a} \;=\;\frac{\text{-}(\text{-}2ah)}{2(a)} \;=\;\frac{2ah}{2a} \;=\;h$

$\text{The vertex occurs at }\,x \,=\, h.$

[2] . $y \:=\:ax^2 - a(p+q)x + apq \quad\Rightarrow\quad \begin{Bmatrix}a &=& a \\ b &=& \text{-}a(p+q) \\ c &=& apq \end{Bmatrix}$

Vertex: . $x \;=\;\frac{\text{-}b}{2a} \;=\;\frac{\text{-}[\text{-}a(p+q)]}{2(a)} \;=\;\frac{a(p+q)}{2a} \;=\;\frac{p+q}{2}$

$\text{The vertex occurs at }\,x \,=\,\tfrac{p+q}{2}$

Thank you so much. That cleared things up but I wanted to know how you went from y=ax^2-apx-aqx+apq to y=ax^-a(p+q)x+apq

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