This is also driving my crazy. I'm asked to write in standard form of the quadratic function. It gives me a picture of a parabola and 3 points, 1 of them is the vertex.

Here's an example:

I don't get it at all.

2. Originally Posted by BeSweeet
This is also driving my crazy. I'm asked to write in standard form of the quadratic function. It gives me a picture of a parabola and 3 points, 1 of them is the vertex.

Here's an example:

I don't get it at all.
Recall that the standard form is $\displaystyle f(x)=a(x-h)^2+k$, where $\displaystyle (h,k)$ is the vertex.

Then plug in any point that is given to you in the graph above to find $\displaystyle a$.

Can you try to take it from here?

3. It wants me to write it in standard form, not solve for a... Or am I missing something?

Here's what the problem looks like:

4. Originally Posted by BeSweeet
It wants me to write it in standard form, not solve for a... Or am I missing something?
you need to solve for "a" to complete the equation.

5. So I can pick one of those two points? Trying it now.

6. Awesome! Definitely have this stuff figured out now. Thanks to everyone who helped with those quick responses!!

7. Ok I'm not trying to hijack this post but got a quick question.

Using Sum and Product of the roots you can get the equation but how would you know which is correct without using the graph

Sum = 0
Product = -1

so

$\displaystyle x^2 -1$ which is incorrect

or

$\displaystyle x^2 + 1$ which is correct

8. Originally Posted by 11rdc11
Ok I'm not trying to hijack this post but got a quick question.

Using Sum and Product of the roots you can get the equation but how would you know which is correct without using the graph

Sum = 0
Product = -1

so

$\displaystyle x^2 -1$ which is incorrect

or

$\displaystyle x^2 + 1$ which is correct
correction ... $\displaystyle 1-x^2$

9. Originally Posted by skeeter
correction ... $\displaystyle 1-x^2$
oops lol thanks. Otherwise I have imaginary numbers. Forgot to put the negative infront of x^2 so the graph opens downward. But skeeter how would I know that $\displaystyle -x^2 +1$ is the graph and not $\displaystyle x^2 -1$ just using the formula $\displaystyle x^2 - Sx + P$

10. Originally Posted by 11rdc11
oops lol thanks. Otherwise I have imaginary numbers. Forgot to put the negative infront of x^2 so the graph opens downward. But skeeter how would I know that $\displaystyle -x^2 +1$ is the graph and not $\displaystyle x^2 -1$ just using the formula $\displaystyle x^2 - Sx + P$
you can't know w/o additional info ... both have the same sum/product of roots.

11. Originally Posted by skeeter
you can't know w/o additional info ... both have the same sum/product of roots.
O ok thanks that is what I thought