whats the equationo of the parabola with vertex (-2,4) that passes through the point (0,1)?

Answer 1

equation of the parabola with vertex(-2,4)
=a(x+2)^2=y-4
passes through (0,1)
plugging in
a(0+2)^2=1-4
4a=-3
a=-3/4
so the equation of the parabolais
-3(x+2)^2=4(y-4)

Answer 2

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

(0,1) gives us f(x) = ax^2 + bx + 1

x = (-b)/(2a)
-2 = (-b)/(2a)
-b = -4a
b = 4a

f(-2) = a(-2)^2 + 4a(-2) + 1
4 = 4a - 8a + 1
4a - 8a = -1 + 4
-4a = 3
a = (-3/4)

b = 4(-3/4) = -3

ANS : f(x) = (-3/4)x^2 - 3x + 1

for proof, go to http://www.calculator.com/calcs/GCalc.html and just copy and past everything after f(x) =

Answer 3

y=-0.75(x + 2)^2 + 4