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$$\text{The parabola with equation y=ax^2+bx+c is graphed below:}$$ $$\text{The zeros of the quadratic ax^2 + bx + c are at x=m and x=n, where m>n. What is m-n? }$$

Jul 18, 2019

#1
+2

The x coordinate of the vertex  can be found as   -b / (2a)  =  2   ⇒  -b = 4a  ⇒  b  = -4a

y = ax^2 - 4ax + c

And since  (2,1)   and (-4, -3)  are on the graph, we have that

a(2)^2 - 4a(2) + c  =  1  ⇒   4a - 8a + c  = 1  ⇒    -4a + c  =  1   ⇒   -32a + 8c  = 8    (1)

a(-4)^2  - 4a(-4) + c  = -3  ⇒   16a + 16a + c  = -3  ⇒  32a + c  = -3      (2)

Add (1)  and (2)   and we have that

9c = 5

c = 5/9

And

-4a + 5/9  =  1

-4a =  4/9

a  = -1//9

And b  = -4(-1/9)  = 4/9

y  = (-1/9)x^2 + (4/9)x + 5/9

Setting this to 0  and we have that

(-1/9)x^2  + (4/9)x + (5/9)  =  0         multiply through by  -9

x^2 - 4x - 5  = 0       factor

(x - 5) ( x + 1)  = 0      set each factor to 0 and solve for x and we have that

x - 5  = 0                x + 1  = 0

x  = 5  = m             x  = -1 = n

So....m - n  =   5 -(-1)   =   6   Jul 18, 2019