# Thread: Writing the vertex of a parabola

1. ## Writing the vertex of a parabola

Just like a second opinion please.

I have a parabola y = x^2 - 10x - 8

I have the equation y = a(x - h)^2 + K

where (h, k) are the vertex

am I right to change the signs and write;

h = 10
k = 8

so the vertex of y = x^2 - 10x - 8

= (10, 8)

Thanks

2. ## Re: Writing the vertex of a parabola

You've to convert $\displaystyle y=x^2-10x-8$ into the form $\displaystyle y=a(x-h)^2+k$.
In you other topic you used the technique 'completing the square', use this technique for this exercice.

Notice:
The vertex $\displaystyle V(10,8)$ is the vertex of the parabola: $\displaystyle y=(x-10)^2+8=x^2-20x+108$.

3. ## Re: Writing the vertex of a parabola

Originally Posted by Siron
You've to convert $\displaystyle y=x^2-10x-8$ into the form $\displaystyle y=a(x-h)^2+k$.
In you other topic you used the technique 'completing the square', use this technique for this exercice.

Notice:
The vertex $\displaystyle V(10,8)$ is the vertex of the parabola: $\displaystyle y=(x-10)^2+8=x^2-20x+108$.
Thanks for that, so if I already have the equation and the parabola, I can just read straight from the parabola and apply it to the h and k, i.e the vertex (10, 8).

4. ## Re: Writing the vertex of a parabola

If you've a polynomial in the form:
$\displaystyle y=a(x-h)^2+k$
then like you said, you can determine the coordinates of the vertex directly: (h,k)

If you've a polynomial in the form:
$\displaystyle y=ax^2+bx+c$
then you've to convert this polynomial into the form $\displaystyle y=a(x-h)^2+k$

In case of your exercice, you have given a polynomial in the form $\displaystyle y=ax^2+bx+c$ so convert it to the other form (by using 'completing the square, like you did in your previous topic)

So, can you write:
$\displaystyle y=x^2-10x-8$ into the form: $\displaystyle y=a(x-h)^2+k$?

5. ## Re: Writing the vertex of a parabola

Originally Posted by Siron
If you've a polynomial in the form:
$\displaystyle y=a(x-h)^2+k$
then like you said, you can determine the coordinates of the vertex directly: (h,k)

If you've a polynomial in the form:
$\displaystyle y=ax^2+bx+c$
then you've to convert this polynomial into the form $\displaystyle y=a(x-h)^2+k$

In case of your exercice, you have given a polynomial in the form $\displaystyle y=ax^2+bx+c$ so convert it to the other form (by using 'completing the square, like you did in your previous topic)

So, can you write:
$\displaystyle y=x^2-10x-8$ into the form: $\displaystyle y=a(x-h)^2+k$?
If I am right, then;

y = x^2 - 10x + 8 = x(x - 10)^2 - 8

6. ## Re: Writing the vertex of a parabola

But $\displaystyle y=x(x-10)^2-8=x(x^2-20x+100)-8=x^3-20x^2+100x-8$, so that's not an option.
Do you notice:
$\displaystyle x^2-10x+25=(x-5)^2$
?
But in this case you've $\displaystyle y=x^2-10x-8$, so you've to subtract a (constant) number of 25 to hold the original equation, so:
$\displaystyle x^2-10x+25-a=x^2-10x-8$

Calculate a, then you can wright:
$\displaystyle x^2-10x-8=(x-5)^2-a$

7. ## Re: Writing the vertex of a parabola

Originally Posted by David Green
I have a parabola y = x^2 - 10x - 8
I have the equation y = a(x - h)^2 + K
where (h, k) are the vertex
$\displaystyle y=x^2-10x-8=(x-5)^2-33$ Completing the square.
So $\displaystyle h=5~\&~k=-33.$

8. ## Re: Writing the vertex of a parabola

Originally Posted by Siron
But $\displaystyle y=x(x-10)^2-8=x(x^2-20x+100)-8=x^3-20x^2+100x-8$, so that's not an option.
Do you notice:
$\displaystyle x^2-10x+25=(x-5)^2$
?
But in this case you've $\displaystyle y=x^2-10x-8$, so you've to subtract a (constant) number of 25 to hold the original equation, so:
$\displaystyle x^2-10x+25-a=x^2-10x-8$

Calculate a, then you can wright:
$\displaystyle x^2-10x-8=(x-5)^2-a$
Thanks Siron, Plato has just beat me to it with the solution, but I have worked it out like this;

y = x^2 - 10x - 8 = (x - 5)^2 - 25 - 8

= (x - 5)^2 - 33

vertex is (5, - 33)

Thanks to you both much appreciated.

9. ## Re: Writing the vertex of a parabola

Well done and you're welcome!

10. ## Re: Writing the vertex of a parabola

Originally Posted by Siron
Well done and you're welcome!
Thanks

David