y=ax^{2} +bx+c

Vertices (-1,-5), (0,-1), (2,1)

Find the standard form of the parabola passing through those points

Guest Jan 11, 2019

edited by
Guest
Jan 11, 2019

edited by Guest Jan 11, 2019

edited by Guest Jan 11, 2019

#1**+1 **

Since we have the point (0, -1) this is the y intercept ....so c = -1

And we have this system of equations

a(-1)^2 + b(-1) - 1 = -5

a(2)^2 + b(2) - 1 = 1

Simplifying these, we have

a - b = - 4 (1)

4a + 2b = 2 ⇒ 2a + b = 1 (2)

Add (1) and (2) and we have that

3a = -3

a = - 1

So a - b = -4

-1 - b = - 4

b = 3

So.....the equation of the parabola is

y = -x^2 + 3x - 1

Here's the graph : https://www.desmos.com/calculator/pgv8twlmd1

CPhill Jan 11, 2019