If it has the vertex (1,-4) and passes through the point (2,3) what is this in quadratic form?

Question

anonymous · Answer

Formula is f(x)=a(x-h)^2+k

et365 · Answer

That is in vertex form, I understand that, but I'm not too sure what it is in quadratic form

anonymous · Answer

Points are (x,y) =(2,3)
Vertex is (h,k) = (1,-4)

3=a(2-(1))^2+(-4)
Do the parentheses first which is 
3=a(1)^2 +(-4)
Then do the exponent 
3=a(1)+(-4)
Add 4 to both sides
7=a
Then use the formula and plug it in 
So the answer should be
F(x)=7(x-1)-4

campbell_st · Answer

well when you find a, but substituting the point you can then distribute to get the equation

anonymous · Answer

This is the standard form for quadratics

campbell_st · Answer

and it should be 

$$f(x) =7(x -1)^2 - 4 $$

anonymous · Answer

I forgot to put the square in sorry

et365 · Answer

Hmm so any way to convert that to non-standard?

campbell_st · Answer

so you want it as

$$f(x)=ax^2 + bx + c$$

is that correct...?

et365 · Answer

Yes

campbell_st · Answer

ok... well its 

$$f(x) = 7(x^2 -2x + 1) - 4$$

just distribute the 7 and collect like terms

et365 · Answer

Thanks, I was unsure of which form of the equation to sub a in.