Application of Quadratics in space exploration part I & II

Question

Part I

Team Johnson will use algebra, just as Katherine Johnson did in real life, to come up with calculations, numbers, and ultimately, a function. You are going to come up with an equation to map out John Glenn’s descent back to Earth in Friendship 7. You are supplied with 2 graphs that have relevant values on both the x and y axis. You are supplied with the following points: (0,160) and (28,20);

The coordinates represent targeted points that we are trying to get Friendship 7 to go through in its parabolic path as John Glenn comes back to Earth. ***hint: assume that (0,160) in the high point in his parabolic (parabola) descent.

PART II

You are going to come up with an equation to map out John Glenn’s descent back to Earth in Friendship 7. You are supplied with 2 graphs that have relevant values on both the x and y axis. You only need to use 1 of them, but there is an extra one in case you wish to use it. You are going to plot the following points on this graph: (0,160), (10,150), (17.5,130), (22,110), (27,70), (30,0);

These coordinates represent approximate targeted points that we are trying to get Friendship 7 to go through in its parabolic (parabola) path as John Glenn comes back to Earth. Come up with a function, a graph of “best fit”, that represents the line that Friendship 7 will travel. Sketch your equation on the same graph as your given plotted points.

Plot the provided points on the axis given. After calculating the equation, sketch your equation on the same graph as your given plotted points.

Answer 1

(0, 160) is the vertex = max point

y = a(x-0)^2 + 160 in vertex form y =a(x-h)^2 +k where (h,k) = vertex = (0, 160)

y=ax^2 +160

plug in the other point (28,20) to solve for a

20 =a(28)^2 + 160

-140 =a28^2

a = -140/28^2

a =-140/784

a =--5/28

y = -5x^2/28 + 160

Check the equation by plugging in the points

20 = -5(28)^2/28 +160

20 =-5(28)+160

= -140+160

160=-5(0)^2/28 +160

= 160

the graph is a downward opening parabola

with vertex (0,160) = maximum point = y intercept

axis of symmetry is x=0 or the y axis