Question

Ok this is a quadratic's question: a parabola has the x intercepts of 0 and 6, determine the expanded form of the algebraic relation that defines this parabola.

Answer 1

Also the y intercept of the vertex is 3 if that helps

Answer 2

Use the intercept form of a parabola:
y=a(x-x1)(x-x2)
where a is a constant to be determined, and x1,x2 are the x-intercepts of the function.

\(Example\):
If the intercepts (roots) are 0,4, and the vertex has y=1.
then the equation of the parabola is
y=a(x-0)(x-4)
=ax(x-4)

The parabola is symmetrical about the vertex, so the vertex is at x=(0+4)/2=2
Substitute known values, y=1, x=2 into the equation
1=a(2-0)(2-4)=a(2)(-2)=-4a
so a=-1/4
and the complete equation is
y=-(1/4)(x)(x-4)

Answer 3

Ok I got -1/3 for a but I don't know how to find the rest of the equation.. @mathmate

Answer 4

-1/3 looks good.
Can you show me how you can get that without knowing the rest of the equation. I am curious.

Answer 5

Ok so I got the x intercept of the vertex by adding them and then dividing them by 2 to get 3, then I plugged in all the numbers I knew into the original equation: y=a(x-s)(x-t) = 3=a(3-0)(3-6) and then I just kept simplifying until I got a :) however I'm very confused about calculating the rest of the equation @mathmate

Answer 6

I added the x intercepts (0 and 6) together and divided by 2 to get the x part of the vertex

Answer 7

ok, you already have the right equation and the right way to do it.
Unfortunately you did it blindly (but correctly).
Please read my example, which is exactly what you did, but for a different problem.
See if you can try to understand the procedure, and it will help you solve other problems.
Please let me know after reading and understanding my example.

Answer 8

Ok I understood your example thanks!

Answer 9

You're welcome! :)