Quadratics Math Help

Question

When Evil Kenevil was a kid, he decided to attempt to jump a distance of 30 yards using ramps that extended 9 yards high for take off and landing with his motorcycle.

As the figure below illustrates, at zero yards from the take off ramp, he was 9 yards high. At 8 yards from the ramp, he was 17 yards high, and at 18 yards from the ramp, he was 18 yards high.

Since there are no rockets or wings affecting Evil’s flight, he should closely follow a parabolic path that can be modeled by a quadratic equation:
y=ax^2+bx+c, where y is the height of Evil’s bike, and x is the horizontal distance he has traveled.





1.) Using the three data points (x, y) that we know Evil passed through and the general quadratic equation, create three corresponding equations:

Equation 1: 


Equation 2: 


Equation 3: 


2.) Solve your system of equations from above for a, b, and c.

a = 


b = 


c = 


3.) Using your values of a, b, and c, what is the quadratic equation that models Evil’s jump?

Quadratic Equation: 

4.) What is the maximum height for Evil during this jump? 

5.) What is Evil’s height when he gets to the landing ramp? Is this a successful stunt? 

6.) Evil was at a height of 17 yards when he was 8 yards out. Where else was he at a height of 17 yards? 

Answer 1

(0,9)

(8,17)

(18,18)

 

From the first given data point, the y-intercept, we get C=9  <---   #1) first equation

 

  (8,17) ---> 64A + 8B + 9 = 17

                  64A + 8B = 8

                 8A + B = 1           <---- 2nd equation

 

 

(18,18)  ---> 324A + 18B + 9 = 18

                     324A + 18B = 9

                     36A + 2B = 1   <---- third equation

 

 

part #2) solving the system of equations

 

36A + 2(1 - 8A) = 1

36A + 2 -16A = 1

20A = -1

A= -1/20

 

8(-1/20) + b = 1

-8/20 +b = 1

 

  b = 1 +8/20

      = 20/20 + 8/20

      = 28/20

     = 14/10

      = 7/5

 

 f(x) = (-1/20)x^2 + (7/5)x + 9  <---- A=-1/20 , B = 7/5, C= 9   : that's part #3

 

 

part 4:

   max occurs = -b/(2a)  = -7/5 divided by (-1/10)  = (-7/5)(-10) = 14

    f(14) = 18.8

 

 

Part 5:

  

 (-1/20)x^2 + (7/5)x + 9 = 0

   x^2 - 28x - 180 = 0 <--- multiplies by -20 to kill the fractions

  

  x = [ 28 + sqrt( (-28)^2 + 4*180)]/2  <--- quadratic formula: negative solution not allowed

 

    = [ 28 + sqrt(1504)]/2

 

    = 28 + sqrt(16*2*47)]/2

 

    = 28 + 4*sqrt(94)]/2

 

   = 14 + 2*sqrt(94)

 

since this is just over 33 yards, yes the jump was successful

 

 

 

part 6)

 

  (-1/20)x^2 + (7/5)x + 9 =17

 

  (-1/20)x^2 + (7/5)x - 8 = 0

 

  x^2 -28x +160 = 0

 (  x   -   20    )(x -   8      ) = 0

 

 x = 20 or x=8 --->  20 yards