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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

#2
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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
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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   Mar 20, 2020