so you have the points:
(-2,1), (0,5), and (4,-59)
let the function be
y = ax^2 + bx + c
for (0,5)
5 = 0 + 0 + c -----> c = 5
so we could reduce our equation to
y = ax^2 + bx + 5
for (-2,1)
1 = 4a - 2b + 5
4a - 2b = -4
2a - b = -2 ***
for (4, -59)
-59 = 16a + 4b + 5
16a + 4b = -64
4a + b = -16 ****
add *** and ****
6a = -18
a = -3
into ****
-12 + b = -16
b = -4
y = -3x^2 - 4x + 5
change to whatever version of the parabola you need
A quick mental check shows that all 3 points lie on the curve
what is the equation in standard form of a parabola that models the value in the table? x -2,0,4 fx 1,5,-59
1 answer