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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
avatar+111387 
+2

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

 

So.....we have the following quadratic

 

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

 

So....the quadratic  is

 

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

 

 

cool cool cool

 Jul 18, 2019

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