Question

write the quadratic function in standard form that has its vertex at 3,-2 and passes through 1,6

Answer 1

what's the vertex?

Answer 2

3,-2*

Answer 3

ok... so the vertex form of a quadratic is

\[y = a(x -h)^2 + k\]

you know h = 3 and k = -2

substitute then to get the equation

next you need to find a

substitute x = 1 and y = 6 and solve for a

Answer 4

Vertex form:

y = a(x-h)^2 + k

we know the vertex is (3, -2), so (h,k) = (3, -2) ---> h = 3 and k = -2

y = a(x-h)^2 + k

y = a(x-3)^2 + (-2)

y = a(x-3)^2 - 2

Now plug in (x,y) = (1,6) or x = 1 and y = 6 to get

y = a(x-3)^2 - 2

6 = a(1-3)^2 - 2

6 = a(-2)^2 - 2

6 = 4a - 2

keep going to solve for 'a'