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

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: $\displaystyle y=(x-10)^2+8=x^2-20x+108$.

Re: Writing the vertex of a parabola

Quote:

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).

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: $\displaystyle y=a(x-h)^2+k$?

Re: Writing the vertex of a parabola

Quote:

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

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:

$\displaystyle x^2-10x-8=(x-5)^2-a$

Re: Writing the vertex of a parabola

Quote:

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 $\displaystyle h=5~\&~k=-33.$

Re: Writing the vertex of a parabola

Quote:

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.

Re: Writing the vertex of a parabola

Well done and you're welcome! :)

Re: Writing the vertex of a parabola

Quote:

Originally Posted by

**Siron** Well done and you're welcome! :)

Thanks

David