the axis of symmetry, the vertex, any x-and y-intercepts and graphing the parabolas.

here's an example question

y=-4x^2-9x-2

also y=x^2-6x+25

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- January 2nd 2010, 11:38 PMVuongNeed help solving for:
the axis of symmetry, the vertex, any x-and y-intercepts and graphing the parabolas.

here's an example question

y=-4x^2-9x-2

also y=x^2-6x+25 - January 2nd 2010, 11:42 PMDrexel28
- January 2nd 2010, 11:47 PMVuong
I never understood what the f(x) meant...

- January 3rd 2010, 12:26 AM11rdc11
- January 3rd 2010, 01:49 AMWilmer
- January 3rd 2010, 01:16 PMVuong
- January 3rd 2010, 02:01 PMQuacky
y=-4x^2-9x-2

Ok, the line will cut the y axis where x=0.

Substitute x=0 into the equation, and you get y=-2. So point of interception is at:

(0,-2)

Now, the line will cut the x axis where y=0. To find this, we can try to factorise:

x=-2 and x=-1/4

x intercepts are

and

Parabolas are symmetrical, so if you take the x intercepts, the line of symmetry will be halfway between them.

(wow, nasty numbers)

x=-1.125 is the line of symmetry.

A question about the vertex: Have you covered differentiation? - January 3rd 2010, 03:25 PMVuong
- January 3rd 2010, 03:30 PMQuacky
(4x+1)(x+2)=0

Substitute x=-2 in and see what happens. - January 3rd 2010, 03:33 PMVuong
- January 3rd 2010, 03:59 PMQuacky
Yes.

If you substitute 2 into the equation, you get

As you want the solutions to

(not 36)

x=-2 is the correct solution, not x=2. - January 3rd 2010, 04:00 PMVuong
- January 3rd 2010, 04:05 PMQuacky
Think of it this way:

http://www.mathhelpforum.com/math-he...505670c8-1.gif

For that to be true, either

or

If

If

Edit: I'll look at the above problem now. - January 3rd 2010, 04:05 PMDrexel28
- January 3rd 2010, 04:10 PMVuong