Formula for Parabola

Question

Find a formula for the quadratic function whose graph has its vertex at (1,8) and its y-intercept at y=−2.

Answer 1

Vertex form for a parabola: y = A(x-h)²+h where (h,k) is the vertex.

So,

y = A(x-1)²+8

When x = 0, y = -2 (y-intercept):

-2 = A(0-1)²+8

-2 = A+8

A=-10

So the function is defined by:

y = -10(x-1)²+8

desmos.com/calculator/6pn9n69rch

Answer 2

y = a(x-1)^2 + 8 = -2 when x= 0

-2 = a +8

a = -10

y = -10(x-1)^2 + 8 has vertex (1,8) and y intercept =-2

or

y = -10x^2 +20x -2

that's a vertical downward opening parabola

for a horizontal parabola,leftward opening

x = a(y-8)^2 +1

use the y intercept to find the value of "a"

let x=0, then (y-8)^2 = 1/a

y-8 = sqr(1/a)

y = +/-sqr(1/a) = -2

-10 = 1/a

a=-1/100

x = -.1(y-8)^2 +1

or

x= -.1y^2 +1.6y -5.4

or

=-y^2/100 +8y/5-27/5

but it has a 2nd y intercept = 18

then there are endless other parabolas with the same vertex and one y intercept =-2, at different angles

but all have two y intercepts not just the one y=-2

and all of them are not functions, just relations of x and y

so go with y=a(x-h)^2+k for vertex =(h,k)=(1,8) and solve for "a" using the point (0,-2)