Please help! Will give medal to best answer!

6. Suppose a parabola has a vertex of (5,-3) and passes through the point (6,1). Write the equation of the parabola in vertex form.
A. y=(x-5)^2-3
B. y=4(x-5)^2-3
C. y=4(x+5)^2-3
D. y=4(x-5)^2+3

Question

Please help! Will give medal to best answer!

6. Suppose a parabola has a vertex of (5,-3) and passes through the point (6,1). Write the equation of the parabola in vertex form.
A. y=(x-5)^2-3
B. y=4(x-5)^2-3
C. y=4(x+5)^2-3
D. y=4(x-5)^2+3

campbell_st · Answer

well knowing the vertex you can write it as

$$y = a(x - h)^2 + k$$

where (h, k) is the vertex. 

so substitute your vertex. 

to find the value of a, substitute the point (6, 1) into the vertex equation to get thew value of a.

then you'll have your equation

dinorap1 · Answer

Oh, I see... so that's how you're supposed to do it. Okay, I'll try it and see what I get.

campbell_st · Answer

glad to help

dinorap1 · Answer

But I've input it into my graphing calculator, and the only answers that might be right are A or B. Is that right or wrong?

dinorap1 · Answer

If I knew how to find certain coordinates, I would already know the answer by now...

campbell_st · Answer

ok... so the vertex form after putting the vertex in is 

$$y =a(x - 5)^2 - 3$$

now substitute x = 6 and y = 1 to find a

dinorap1 · Answer

So when you plug that in, it becomes:
1=a(6-5)^2-3

Do you distribute a first, or do you distribute the ^2 first?

campbell_st · Answer

no just go 

$$1 = a \times (1)^2 -3$$

whcih becomes

1 = a - 3

solve for a

dinorap1 · Answer

Sorry, I get confused whenever there's an exponent hanging around...
Okay, that makes sense. So:
a=4

campbell_st · Answer

thats correct so you should eb able to identify the correct equation now

dinorap1 · Answer

It's B, right? 
Thanks for the help by the way! :)

campbell_st · Answer

that's correct