• Mar 9th 2010, 11:28 PM
Anemori
Ok here some excercise from the book that I don't get it because the book the does not have an example how to do this.

f(x)= .5x^2-.5

I couldn't figure out how to transform this to standard form of quadratic function from general form.

please anyone can explain how.... thanks!
• Mar 9th 2010, 11:32 PM
Prove It
Quote:

Originally Posted by Anemori
Ok here some excercise from the book that I don't get it because the book the does not have an example how to do this.

f(x)= .5x^2-.5

I couldn't figure out how to transform this to standard form of quadratic function from general form.

please anyone can explain how.... thanks!

Take out the common factor of $0.5$.

$f(x) = 0.5(x^2 - 1)$

$= 0.5[(x - 0)^2 - 1]$

$= 0.5(x - 0)^2 - 0.5$
• Mar 10th 2010, 12:01 AM
Anemori
Quote:

Originally Posted by Prove It
Take out the common factor of $0.5$.

$f(x) = 0.5(x^2 - 1)$

$= 0.5[(x - 0)^2 - 1]$

$= 0.5(x - 0)^2 - 0.5$

Thanks. I'm right with my guess then hehehe...

what I did is that I did not transform it to standard form.

f(x) = .5x^2-.5

a= .5 , b= 0 , c= .5

x= -b/2(a) ==> 0/2(.5) =0

x= 0

f(0) = .5(0)^2-.5 = -.5

y = -.5

vertex point of the graph is (0, -.5)
• Mar 10th 2010, 02:59 AM
HallsofIvy
That should have been obvious immediately.

Since a square is never negative, $.5x^2$ is never negative so $y= .5+ .5x^2$ is ".5 plus something". That is, y is never less than .5 and will be equal to .5 only when x= 0. The vertex, the lowest point on the graph, is at (0, .5).

(Don't just memorize formulas. Think about what you are doing!)