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# help

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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 is bigger than n. What is m-n?

Jan 6, 2019

#1
+4

Here's the equation of the curve: Can you take it from here?

Jan 6, 2019
#2
+1

Here's another way

In the form   ax^2 + bx + c, the equation of the x coordinate of the vertex is given by   -b/ 2a

So.....   2 =  -b / [2a]  ⇒   4a = - b  ⇒  b = - 4a

And we know that

1 = a(2)^2 - 4a(2) + c           and    -3  =  a(-4)^2 - 4a(-4)  + c

1  = 4a - 8a + c                              -3 = 16a + 16a + c

1 = -4a + c                                      -3 = 32a + c    ⇒    3 = -32a - c

Add the last two equations and we have that

4 = -36a

-4/36 = a

-1/9 = a

And   b = -4a =  -4(-1/9) = 4/9

To find c, we have

1 = (-1/9)(2)^2 + (4/9)(2) + c

1 =  (-4/9) + 8/9 + c

1 = 4/9 + c

5/9 = c

So.....the function is

y = (-1/9)x^2 + (4/9)x + 5/9

To find tthe zeroes, set this = 0

(-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

Setting each factor to  0 and solving for x and we get  that   x = - 1  or x = 5

So   m = 5  and n = -1

So

m - n =   5 -  (-1)   =   6   Jan 6, 2019