# Quadratic function:convert to standard form?

• September 1st 2012, 12:03 PM
Husky93
Hi guys, i need help with these questions, i have attempted the questions myself and just wanted to double check my answers.
convert to standard form
1. y = –3x2 + 2x + 1 ( the answer i got is Y=-3(x+1/3)^2 -2/3) is this correct?
2.
y = 4(x + 5)(x – 2) ( i can't figure how to solve this one)
3. Y= -1/2 x^2 ( the answer i got is Y= 2(1-3)+2
can someone please chck my answers and correct me i was wrong.
• September 1st 2012, 01:14 PM
MaxJasper
Re: Quadratic function:convert to standard form?
I suspect standard form means f(x)=a*x^2 + b*x +c
• September 1st 2012, 03:47 PM
Soroban
Re: Quadratic function:convert to standard form?
Hello, Husky93!

This is not Standard Form.

You want the "Vertex Form": . $y \:=\:a(x-h)^2 + k$
. . where the vertex is $(h,k).$

Quote:

Convert to Vertex Form.

$(1)\;y \:=\:–3x^2 + 2x + 1$

$y \;=\;-3\left(x^2 - \tfrac{2}{3}x - \tfrac{1}{3}\right)$

$y \;=\;-3\left(x^2 - \tfrac{2}{3}x \:{\color{red}+\: \tfrac{1}{9}} - \tfrac{1}{3} \:{\color{red}-\: \tfrac{1}{9}}\right)$

$y \;=\;-3\left(\left[x - \tfrac{1}{3}\right]^2 - \tfrac{4}{9}\right)$

$y \;=\;-3\left(x-\tfrac{1}{3}\right)^2 + \tfrac{4}{3}$

$\text{Vertex: }\: \left(\tfrac{1}{3},\:\tfrac{4}{3}\right)$

Quote:

$(2)\;y \:=\: 4(x + 5)(x – 2)$

$y \;=\;4(x^2 + 3x - 10)$

$y \;=\;4\left(x^2 + 3x\:{\color{red}+\:\tfrac{9}{4}} - 10\:{\color{red}-\:\tfrac{9}{4}}\right)$

$y \;=\;4\left(\left[x+\tfrac{3}{2}\right]^2 - \tfrac{49}{4}\right)$

$y \;=\; 4\left(x + \tfrac{3}{2}\right)^2 - 49$

$\text{Vertex: }\:\left(\text{-}\tfrac{3}{2},\:\text{-}49\right)$

Quote:

$(3)\; y\:=\: -\tfrac{1}{2}x^2$

This is already in vertex form: . $y \;=\;-\tfrac{1}{2}(x - 0)^2 + 0$

$\text{Vertex: }\:(0,\,0)$
• September 1st 2012, 08:29 PM
Prove It
Re: Quadratic function:convert to standard form?
Quote:

Originally Posted by Soroban
Hello, Husky93!

This is not Standard Form.

You want the "Vertex Form": . $y \:=\:a(x-h)^2 + k$
. . where the vertex is $(h,k).$

$y \;=\;-3\left(x^2 - \tfrac{2}{3}x - \tfrac{1}{3}\right)$

$y \;=\;-3\left(x^2 - \tfrac{2}{3}x \:{\color{red}+\: \tfrac{1}{9}} - \tfrac{1}{3} \:{\color{red}-\: \tfrac{1}{9}}\right)$

$y \;=\;-3\left(\left[x - \tfrac{1}{3}\right]^2 - \tfrac{4}{9}\right)$

$y \;=\;-3\left(x-\tfrac{1}{3}\right)^2 + \tfrac{4}{3}$

$\text{Vertex: }\: \left(\tfrac{1}{3},\:\tfrac{4}{3}\right)$

$y \;=\;4(x^2 + 3x - 10)$

$y \;=\;4\left(x^2 + 3x\:{\color{red}+\:\tfrac{9}{4}} - 10\:{\color{red}-\:\tfrac{9}{4}}\right)$

$y \;=\;4\left(\left[x+\tfrac{3}{2}\right]^2 - \tfrac{49}{4}\right)$

$y \;=\; 4\left(x + \tfrac{3}{2}\right)^2 - 49$

$\text{Vertex: }\:\left(\text{-}\tfrac{3}{2},\:\text{-}49\right)$

This is already in vertex form: . $y \;=\;-\tfrac{1}{2}(x - 0)^2 + 0$

$\text{Vertex: }\:(0,\,0)$

The turning point form IS often referred as the Standard form, because it gives the most information.