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

Printable View

- Jan 2nd 2010, 10:38 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 - Jan 2nd 2010, 10:42 PMDrexel28
Let $\displaystyle f(x)=ax^2+bx+c$. A little completing the square shows that $\displaystyle f(x)=a\left(a+\tfrac{b}{2a}\right)^2+c-\frac{b^2}{4a}$. From this we see that the vertex is at $\displaystyle \left(\tfrac{-b}{2a},c-\tfrac{b^2}{4a}\right)$, the axis of symmetry at $\displaystyle x=\frac{-b}{2a}$, the x-intercepts are at $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ given $\displaystyle \Delta\geqslant0$, and the y intercepts are $\displaystyle f(0)=c$.

- Jan 2nd 2010, 10:47 PMVuong
I never understood what the f(x) meant...

- Jan 2nd 2010, 11:26 PM11rdc11
- Jan 3rd 2010, 12:49 AMWilmer
- Jan 3rd 2010, 12:16 PMVuong
- Jan 3rd 2010, 01:01 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:

$\displaystyle -4x^2-9x-2=0$

$\displaystyle 4x^2+9x+2=0$

$\displaystyle (4x+1)(x+2)=0$

x=-2 and x=-1/4

x intercepts are

$\displaystyle (-2,0)$ and $\displaystyle (-1/4,0)$

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

$\displaystyle -2.25/2 = -1.125$ (wow, nasty numbers)

x=-1.125 is the line of symmetry.

A question about the vertex: Have you covered differentiation? - Jan 3rd 2010, 02:25 PMVuong
- Jan 3rd 2010, 02:30 PMQuacky
(4x+1)(x+2)=0

Substitute x=-2 in and see what happens. - Jan 3rd 2010, 02:33 PMVuong
- Jan 3rd 2010, 02:59 PMQuacky
Yes.

If you substitute 2 into the equation, you get

$\displaystyle (9)(4) = 36.$

As you want the solutions to

$\displaystyle (4x+1)(x+2)=0$ (not 36)

x=-2 is the correct solution, not x=2. - Jan 3rd 2010, 03:00 PMVuong
- Jan 3rd 2010, 03:05 PMQuacky
Think of it this way:

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

For that to be true, either

$\displaystyle (4x+1)=0 $ or $\displaystyle (x+2)=0 $

If $\displaystyle 4x+1=0$

$\displaystyle 4x=-1$

$\displaystyle x=-1/4$

If $\displaystyle x+2=0$

$\displaystyle x=-2$

Edit: I'll look at the above problem now. - Jan 3rd 2010, 03:05 PMDrexel28
- Jan 3rd 2010, 03:10 PMVuong