y=ax2 +bx+c
Vertices (-1,-5), (0,-1), (2,1)
Find the standard form of the parabola passing through those points
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