Question

Help please!
How do you convert standard form into general, vertex form?

Answer 1

...Well i assume you are talking about a quadratic right VTM? lol

They have three forms:
\[ax^2+bx+c\to a(x-h)^2+k\to a(x-r)(x-s)\]

Answer 2

yes i am talking about a quadratic sorry. could you explain the process to me?

Answer 3

Yeah sure so let's take the beautiful example of the quadratic with a vertex at the point \((2,-9)\)? Lol

Answer 4

I'll also tell you our a-value is 1 haha

Answer 5

You have the vertex form like so:
\[\eqalign{
&y=a(x-h)^2+k \\
&y=(1)(x-2)^2-9 \\
&y=(x-2)^2-9 \\
&y=x^2-4x+4-9 \\
&y=x^2-4x-5 \\
}\]
So we started out with vertex form and ended up with standard form by expanding the thing out.

Answer 6

So now let's take a look at going from standard to vertex! We complete the square:
\[\eqalign{
&y=x^2-4x-5 \\
&y=(x^2-4x)-5 \\
&y=(x^2-4x+4-4)-5 \\
&y=(x^2-4x+4)-4-5 \\
&y=(x-2)^2-9
}\]
So we are back where we started. Cool?

Answer 7

yea but i didnt have time to type you that the equation I need to convert is f(x) = x^2 - 10x + 24 to general, vertex from..... so sorry

Answer 8

PSHH Haha im outta here ;)

No that's fine:
\[x^2-10x+24=(x-6)(x-4)\]

And:
\[\eqalign{
x^2-10x+24&=(x^2-10x)+24 \\
&=(x^2-10x+25-25)+24 \\
&=(x^2-10x+25)-25+24 \\
&=(x-5)^2-1
}\]

Answer 9

Thank you!

Answer 10

No prob vtm :)