y=x^2-x-12
Convert to vertex form.

Question

y=x^2-x-12
Convert to vertex form.

ria23 · Answer

I got y-13=(x-1)^2
;-; I don't think it's right...

anonymous · Answer

Well lets examine both Standard and Vertex form

Standard
$$f(x) = ax^2 + bx + c$$
$$f(x) = a(x - h) + k$$

(h,k) is the vertex so from standard form we can find the x-value of the vertex and then the y-value to find (h,k) and plug those into the first equation and a = 1

The vertex in standard form is 

$$x = \frac{ -b }{ 2a }$$

Does this make sense so far?

anonymous · Answer

woops let me rewrite vertex form

$$f(x) = a(x-h)^{2} + k$$

ria23 · Answer

Yea, that makes sense.

ria23 · Answer

I thought vertex form was written$$y-y _{1}=m(x-x _{1})^{2}$$

anonymous · Answer

Nah you are combining two different forms point-slop form for linear equations with vertex form

anonymous · Answer

so continuing on...

our x-value of the vertex would be

$$\frac{ -b }{ 2a } = \frac{ -(-1) }{ 2 }$$

so our x-value is 1/2, now we got to plug that back into the original

$$f(.5) = (.5)^2 - .5 - 12$$
$$f(.5) = \frac{ 1 }{ 4 } - \frac{ 1 }{ 2 } - 12$$

Get common denominators

$$f(.5) = \frac{ 3 }{ 12 } - \frac{ 6 }{ 12 } - \frac{ 144 }{ 12 }$$
$$f(.5) = \frac{ -147 }{ 12 }$$

So our vertex is $$(\frac{ 1 }{ 2 }, \frac{ -147 }{ 12 })$$

Now putting it into vertex form 

$$y = 1(x-\frac{ 1 }{ 2 })^{2} - \frac{ 147 }{ 12 }$$

ria23 · Answer

Thank yhu for explaining everything. c:

anonymous · Answer

Yep!