Wednesday
April 23, 2014

Homework Help: maths

Posted by Emma on Friday, May 25, 2007 at 4:04am.

how do i find out the the equation of the line of symmetry for
f(x)=ax^2+bx+c ?
please explain it do me thanks!

The line of symmetry runs through the vertex.
so the x of the vertex is -b/(2a)

and the equation of the line of symmetry is x = -b/(2a)

e.g. for f(x) = 2x^2 - 12x + 3
after completing the square you can find the vertex to be (3,-15)

or

the x of the vertex is -(-12)/4 = 3
then f(3) = -15.

the line of symmetry is x = 3

Quote:
The line of symmetry runs through the vertex.
so the x of the vertex is -b/(2a)
and the equation of the line of symmetry is x = -b/(2a)

why is the equation of symmetry for f(x)=ax^2+bx+c is x=-b/2a? i still don't get this...

I suspect you need to use it a couple of times, and it will sink in .

In the meantime, Memorize: for the standard quadratic, the line of symettry is -b/2a.

If you have memorized the quadratic equation, you already have it. Remember the first term in the quadratic equation?
-b/2a

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