1. ## quadratic function standard form

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!

2. 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 $0.5$.

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

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

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

3. 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)

4. 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!)