Thread: 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

You've to convert into the form .
In you other topic you used the technique 'completing the square', use this technique for this exercice.

Notice:
The vertex is the vertex of the parabola: .

Originally Posted by Siron

You've to convert into the form .
In you other topic you used the technique 'completing the square', use this technique for this exercice.

Notice:
The vertex is the vertex of the parabola: .

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).

If you've a polynomial in the form:

then like you said, you can determine the coordinates of the vertex directly: (h,k)

If you've a polynomial in the form:

then you've to convert this polynomial into the form

In case of your exercice, you have given a polynomial in the form so convert it to the other form (by using 'completing the square, like you did in your previous topic)

So, can you write:
into the form: ?

Originally Posted by Siron

If you've a polynomial in the form:

then like you said, you can determine the coordinates of the vertex directly: (h,k)

If you've a polynomial in the form:

then you've to convert this polynomial into the form

In case of your exercice, you have given a polynomial in the form so convert it to the other form (by using 'completing the square, like you did in your previous topic)

So, can you write:
into the form: ?

If I am right, then;

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

But , so that's not an option.
Do you notice:

?
But in this case you've , so you've to subtract a (constant) number of 25 to hold the original equation, so:


Calculate a, then you can wright:

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

Completing the square.
So

Originally Posted by Siron

But , so that's not an option.
Do you notice:

?
But in this case you've , so you've to subtract a (constant) number of 25 to hold the original equation, so:


Calculate a, then you can wright:

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.

Well done and you're welcome!

Originally Posted by Siron

Well done and you're welcome!

Thanks

David