# Thread: Need help solving for:

1. ## Need 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

2. Originally Posted by Vuong
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
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 $\displaystyle f(0)=c$.

3. I never understood what the f(x) meant...

4. Originally Posted by Vuong
I never understood what the f(x) meant...
Same thing as y

5. Originally Posted by Vuong
y=-4x^2-9x-2
Start by factoring -4x^2 - 9x - 2

6. Originally Posted by Wilmer
Start by factoring -4x^2 - 9x - 2
What do I do after?

7. 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?

8. Originally Posted by Quacky
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:

-4x^2-9x-2=0
4x^2+9x+2=0
(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?
Why are they minus?

9. (4x+1)(x+2)=0
Substitute x=-2 in and see what happens.

10. Originally Posted by Quacky
(4x+1)(x+2)=0
Substitute x=-2 in and see what happens.
Uhh (-7)(0) = 0?

11. 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.

12. Originally Posted by Quacky
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.
Great, thanks! Big help. Do you know how to factor a problem like 6x^2-5x-6? I've been trying a bunch of combinations, still unsuccessful

13. Think of it this way:

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.

14. Originally Posted by Vuong
Great, thanks! Big help. Do you know how to factor a problem like 6x^2-5x-6? I've been trying a bunch of combinations, still unsuccessful
$\displaystyle 6x^2-5x-6=0\implies x=\frac{5\pm\sqrt{25-4(6)(-6)}}{12}={{{\tfrac{18}{12}}\brace{\tfrac{-8}{12}}}}$ so that $\displaystyle 6x^2-5x-6=\left(x-\tfrac{18}{12}\right)\cdot\left(x+\tfrac{8}{12}\ri ght)$

15. Originally Posted by Drexel28
$\displaystyle 6x^2-5x-6=0\implies x=\frac{5\pm\sqrt{25-4(6)(-6)}}{12}={{{\tfrac{18}{12}}\brace{\tfrac{-8}{12}}}}$ so that $\displaystyle 6x^2-5x-6=\left(x-\tfrac{18}{12}\right)\cdot\left(x+\tfrac{8}{12}\ri ght)$
You could use quadratic to factor?

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