please please help by listing steps to solve this!!

Question

The function h(x)= -0.2(x-6)^2+1.8 models a boy’s jump into a swimming pool, where x is the horizontal

distance traveled (in feet) and is the height (in feet) above the water. When the boy jumps from a higher

platform, he lands 5 feet farther away. Write a function that models the second jump.

Answer 1

I would substitute (x+5) for x in the original function. It would be h(x) = -0.2(x+5 -6)^2 +1.8 or

h(x) = -0.2(x-1)^2 + 1.8. I did this because the x is the distance horizontally and if he lands 5 feet further then it would be x+5.

Answer 2

Hi Abigail,

This questions is focused on the properties of a parabola. The standard form of a parabola is y = a(x-h)^2 + k where (h,k) is the vertex. The sign of the a factor determines whether the function opens up or down, while the value determines the width of the parabola.

The k factor effectively controls how high (or low) the point of the parabola is, as it is the vertex. This k constant is the number that changes when he jumps from a higher position. However, we don't know where he landed, so we don't know what would be 5 ft in front.

A property we can now exploit is the fact that where the parabola touches the x axis (aka the ground in this question), the y value of the function will be zero. We can use this to find out where he lands:

solve y = 0:

0 = -0.2(x-6)^2 + 1.8

1.8/0.2 = (x-6)^2

+/- sqrt(9) = x-6

+/- 3 + 6 = x

x = 3,9

So if he jumps forward, he lands at x = 9.

Since we want to find the k when this landing point is 5 greater, so x=14, we can say that y(14)=0, so that he hits the ground at zero.

y = -0.2(x-6)^2 + k

0 = -0.2(14-6)^2 + k

0 = -12.8 + k

k = 12.8

so the final answer is y = -0.2(x-6)^2 + 12.8

I encourage you to graph these functions in desmos, and test what each part of the equation changes, and imagine the lines as the path of someone actually jumping.

I hope this is helpful!