# Thread: This parabola question baffles me!

1. ## This parabola question baffles me!

A very good day to everyone here.

I would like you to help me solve this question:

What I don't understand:
Candidly, I don't know how to do this whole question. All I know is that the quadratic formula is $y=ax^2+bx+c$. But how can this formula be applied in finding $p$?

2. You know point A is (0,8)

Substitute x = 0, y = 8, into the equation of the graph, y = (x+p)(2-x) and solve for p. Once you have p, you can solve the rest.

3. Hello Educated, here's my working for (a)

$
y = (x+p)(2-x)$

$8=(0+p)(2-0)$

$8= 2p$

$p= 4$

What are the coordinates of point B?
The question doesn't state any value for x nor y. It's so tough.

4. Now that you have the equation, do you know how to find the roots of an equation? To make y = 0?

$y = (x+4)(2-x)$

If x = 2, y = 0. This one is given for you, it's point c.
If x = ..., y = 0. This one you can solve yourself.

Solve the second root and that's the x-coordinates for b.

5. Oh, I see, so I'll take $y$ as $0$, hence-
$
0 = (x+4)(2-x)$

$x+4 =0$ or $2-x+0$

$x=-4$ or $x=2$

Now, since the $x$ value must be negative at the point $B$,
I should reject $x=2$ and take $x=-4$ as the answer.

Also, the $y$ value at point $B$ must be $0$.

Hence B is (-4,0).

Now I would have to do part (c).

The line of symmetry is the midpoint of the parabola.

What am I suppose to do now?

6. Midpoint of the parabola is halfway between the 2 roots.

7. So is the line of symmetry equals to :

$\dfrac{-4+2}{2}$

$=-1$ ?

8. Yep. Remember that it's only the vertical line that can make it symmetrical.

9. Thank you so much for your help, Educated! My problem is finally solved.