# Find The Quadratic Function That Has The Vertex (4,5) and Whose Graph Passes Through The Point (0,4)Is There A Formula That You Can Plug This Into To Find The Quadratic Function?

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Let f(x) = ax^2 + bx + c

The point (0,4) is on the curve, then:

f(0) = C = 4

**==> c= 4**

The vertex V = (4,5)

but we know that:

xV = -b/2a = 4

==> -b = 8a

==> b = -8a.......(1)

yV = - (b^2 - 4ac)/4a = 5

= - (b^2 - 4*4*a)/4a = 5

==> 16a - b^2 = 20a ........(2)

Substitute (1) in (2):

==> 16a + 64a^2 = 20a

==> -4a + 64a^2 = 0

Factor -4a(1 -16a^2) = 0

==> a= 0 or **a= +-1/4**

==> b = -8a = +- 8/4 = +-2

**==> b = +-2**

Then the function is:

**f(x) = (1/4)x^2 -2x + 4**

**AND**

**f(x) = -(1/4)x^2 +2x + 4**

The equation of the parabola 4a(y -k )= (x-h)^2 be the second segree parbola with vertex at (h,k) with focal length a from (h,k) along the line k.

Since the given vertex is at (4,5), we substitite (h,k) with (4,5).

So the equation of parabola becomes:

4a(y-5) = (x-4)^2.

Since this parabola passes through a point (0 ,4)

4a(4-5) = (0-4)^2.

-4a = 16

a = 16/-4 = -4.

Therfore the equation of the parabola becomes:

4(-4)(y-5) = (x-4)^2

(x-4)^2 +16(y-5) = 0 is the required parabola.