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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 $m-n$?

https://latex.artofproblemsolving.com/0/7/6/07675a24920f72edd65e30b6b78914ddf70403be.png

 Mar 19, 2020
 #1
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cpilllllll

 Mar 20, 2020
 #2
avatar+4569 
+1

Rewrite the quadratic equation as a(x-h)^2+k, where h and k are parts of the vertex.

 

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

 

Since (4,12) is a point on the parabola, we can substitute these values in the equation.

 

12=a(4-2)^2-4, 12=4a-4, 4a=16, a=4

 

Now, the equation is complete: \(y=4(x-2)^2-4, y=4x^2-16x+12\)

 

We then plug y=0, to get x=3, x=1.

 

Thus, m-n=3-1=2.

 Mar 20, 2020
 #3
avatar+111360 
+2

The vertex  is at  ( 2,-4)

 

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

 

We  have  the  form

 

4p ( y - k)  = (x - h)^2      where   (h,k) is the vertex  and (x, y) is a point on the parabola

 

So we have

 

4p (12 - - 4) = ( 4 - 2)^2      simplify

 

4p (16)  = 2^2

 

64p  = 4

 

p  =  4/64  =  1/16

 

We can rewrite  this as

 

4 (1/16) ( y + 4)  =  ( x - 2)^2

 

 (1/4) ( y + 4) = (x - 2)^2       multiply  both sides by 4

 

y + 4  = 4(x - 2)^2     simplify

 

y  = 4 (x^2 - 4x + 4)  - 4

 

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

 

y = 4x^2 - 16x  + 12

 

To find the roots....let  y  = 0

 

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

 

x^2  - 4x  + 3   =  0       factor

 

(x - 3) (x - 1)  =  0

 

Setting both factors to  0   and solving for  x  produces  the roots  x  = 3   and  x  = 1

 

So

 

m  = 3    and  n  =  1

 

So

 

m - n  =   2

 

 

cool cool cool

 Mar 20, 2020

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