What is the vertex form of y= 2x^2 + 9x - 5 ?

1 Answer
Mar 29, 2016

y= 2(x+9/4)^2-121/8

Explanation:

Given:" "y=2x^2+9x-5......................(1)

Write as:" "y=2(x^(color(magenta)(2))+9/2x)-5+k

Where k is a correction factor for an unfortunate consequence of what we are about to do.

Take the power of 2 from x^2 and move it to outside the brackets

" "y=2(x+9/2color(blue)(x))^(color(magenta)(2))-5+k

'Get rid' of the color(blue)(x) from 9/2color(blue)(x)

" "y=2(x+9/2)^2-5+k

Apply (-1/2)xx9/2 = -9/4

" "y=2(x+9/4)^2-5+k .....................................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The error comes from the 9/4 being squared. It introduces an extra value that we not there before. By the way, do not forget to multiply it by the constant of 2 outside the bracket.

So the error is 2(9/4)^2

Consequently it has to be the case that: 2(9/4)^2+k=0

so we have 2(81/16) +k=0

=> k= -81/8

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So equation (2) becomes

" "y=2(x+9/4)^2-5-81/8 .....................................(2_a)

Giving:

color(blue)(" "y= 2(x+9/4)^2-121/8)