What is the equation, in vertex form, of the quadratic function that has a vertex at V(4, 5) and passes through point P(8, 21)

Question

need help

Answer 1

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

21-5 =a(8-4)^2 = 16a

a = 16/16 = 1

y-5 =(x-4)^2

y = (x-4)^2 +5 is in vertex form

for an upward opening parabola

or

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

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

8-4 =a(21-5)^2

4 = a(16)^2

a = 4/16^2 = 1/64

x-4 = (1/64)(y-5)^2

x = (1/64)(y-5)^2 +4 in vertex form for a rightward opening horizontal parabola

8 = (1/64)(21-5)^2+4 = (1/64)(16^2) +4= 256/64 + 4 = 4+4 = 8

there are also an infinite number of slanted or "tilted" parabols with that same vertex through the given point

with xy, terms, as well as x^2, y^2, x or y terms, where the directrix is "tilted," not a vertical or horizontal line.

You probably want the first solution, which is the common standard parabola, upward opening.