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y=ax2 +bx+c

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

Find the standard form of the parabola passing through those points

 
 Jan 11, 2019
edited by Guest  Jan 11, 2019
edited by Guest  Jan 11, 2019
 #1
avatar+94300 
+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

 

 

cool cool cool

 
 Jan 11, 2019

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