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The parabola with equation \(y=ax^2+bx+c\)  is graphed below:



The zeros of the quadratic \(ax^2 + bx + c\)  are at x=m and x=n , where m>n . What is m-n ?

tertre  Mar 12, 2017

Best Answer 

 #2
avatar+2765 
+5

Thanks so much! You're smart.

tertre  Mar 12, 2017
 #1
avatar+87334 
+5

We know that

 

y = a(x - 2)^2 - 4

 

And the point (4,12)  is on the graph....so...

 

12 = a(4 - 2)^2 - 4

12 = 4a - 4

16 = 4a

4 = a

 

So  we have that

 

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

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

y = 4x^2 - 16x + 16 - 4

y = 4x^2 - 16x + 12

 

And to find the roots

 

0 = 4x^2 - 16x + 12      divide through by 4

0 = x^2 - 4x + 3   factor

0 = (x - 3) ( x - 1)      set each factor to 0  and m = 3  and n = 1

 

So....m - n  =  3 - 1 = 2

 

Here's a graph :  https://www.desmos.com/calculator/yvxfjhlbkv

 

 

cool cool cool

CPhill  Mar 12, 2017
 #2
avatar+2765 
+5
Best Answer

Thanks so much! You're smart.

tertre  Mar 12, 2017

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