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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?

 Nov 15, 2019

Best Answer 

 #1
avatar+8891 
+1

Here's one way to solve this:

 

The equation of this parabola in vertex form is

 

y  =  a(x - 2)2  -  4     , where  a  is some constant that "stretches" or "shrinks" the parabola.

 

To find  a  we can use the fact that the parabola passes through the point  (4, 12)

So we know the following must be true:

 

12  =  a(4 - 2)2  -  4

                                Simplify the right side of the equation....

12  =  4a  -  4

                                Add  4  to both sides of the equation.

16  =  4a

                                Divide both sides of the equation by  4

4  =  a

 

So the equation of this parabola is

 

y  =  4(x - 2)2 - 4

 

To find the zeros, we set  y  equal to  0  and solve for  x

 

0  =  4(x - 2)2 - 4

                              Add  4  to both sides of the equation.

4  =  4(x - 2)2

                              Divide both sides of the equation by  4

1  =  (x - 2)2

                              Take the ± square root of both sides.

±1  =  x - 2

                              Split into two equations and solve for  x ...

 

x - 2  =  1       or       x - 2  =  -1

  x  =  3                         x  =  1

 

So the zeros of the quadratic are at   x = 3   and   x = 1

 

3 - 1  =  2

 

Check:   https://www.desmos.com/calculator/wsmg9lgmsw

 Nov 15, 2019
 #1
avatar+8891 
+1
Best Answer

Here's one way to solve this:

 

The equation of this parabola in vertex form is

 

y  =  a(x - 2)2  -  4     , where  a  is some constant that "stretches" or "shrinks" the parabola.

 

To find  a  we can use the fact that the parabola passes through the point  (4, 12)

So we know the following must be true:

 

12  =  a(4 - 2)2  -  4

                                Simplify the right side of the equation....

12  =  4a  -  4

                                Add  4  to both sides of the equation.

16  =  4a

                                Divide both sides of the equation by  4

4  =  a

 

So the equation of this parabola is

 

y  =  4(x - 2)2 - 4

 

To find the zeros, we set  y  equal to  0  and solve for  x

 

0  =  4(x - 2)2 - 4

                              Add  4  to both sides of the equation.

4  =  4(x - 2)2

                              Divide both sides of the equation by  4

1  =  (x - 2)2

                              Take the ± square root of both sides.

±1  =  x - 2

                              Split into two equations and solve for  x ...

 

x - 2  =  1       or       x - 2  =  -1

  x  =  3                         x  =  1

 

So the zeros of the quadratic are at   x = 3   and   x = 1

 

3 - 1  =  2

 

Check:   https://www.desmos.com/calculator/wsmg9lgmsw

hectictar Nov 15, 2019

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