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
+4622 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.
+129899 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
