Quadratic functions and their vertices?

Question

Hello. I'm having a difficult time trying to figure this problem out about quadratic functions and their vertices. The problem says to identify the quadratic function using the vertex of the quadratic function and a point in which the function passes through. I'm assuming I have to use the standard form of a quadratic function. The vertex is (4, -1) and the point is (2, 3).

I would very much like if someone were to show me how to do it step by step instead of just giving me the answer. I don't want to keep posting questions on twenty problems about the same thing.

Not to sound demeaning, but I would appreciate it greatly if correct answers/steps are given. I have seen many times where people would have four different answers to a simple algebraic problem because of a adding/subtracting mistake.

Thank you!

Answer 1

f(x) = ax² + bx + c
vertex(4, -1) says
4 = -b/2a
so
b = -8a..........(1)
it also says that
f(4) = -1
so
16a + 4b + c = -1........(2)
The point (2, 3) says
f(2) = 3
so
4a + 2b + c = 3............(3)
you need to solve equations (1), (2) & (3) simultaneously for a,b & c
subtract (3) from (2) gives
12a + 2b = -4
substitute (1) gives
12a + 2(-8a) = -4
so
a = 1
gives
b = -8a = -8
rearrange (3) gives
c = 3 - 4a - 2b = 3 - 4 - (-16) = 15

∴ f(x) = x² - 8x + 15

,,,.,.,.,.,

Answer 2

If the vertex is (p,q) the standard form of the function is
y=a(x-p)^2+q and you get "a" by substituting the coordinates of
the given point in this equation
y=a(x-4)-1
3=a(2-4)-1 so 4=-2a and a= -2 and y=-2(x-4)^2-1