Question

the function y=________ models the data.how long is the object in the air? round your answer to the nearest hundredth.

Answer 1

Given the table represents a quadratic relation between x and y:

The general form of the relation will be:

`y=a(x-h)^2+k`

we need to find the value of a, h and k

Using the points from the table:

`\begin{gathered} x=0\rightarrow y=150 \\ 150=a\cdot h^2+k\rightarrow(1) \\ \\ x=1\rightarrow y=134 \\ 134=a\cdot(1-h)^2+k\rightarrow(2) \\ \\ x=2\rightarrow y=86 \\ 86=a(2-h)^2+k\rightarrow(3) \end{gathered}`

by solving 1, 2 and 3:

`\begin{gathered} (2)-(3) \\ 134-86=a(1-h)^2-a(2-h)^2 \\ 48=a(1-2h+h^2)-a(4-4h+h^2) \\ 48=a(1-2h+h^2-4+4h-h^2) \\ 48=a(-3+2h)\rightarrow(4) \\ \\ (1)-(3) \\ 150-86=a\cdot h^2-a(2-h)^2 \\ 64=ah^2-a(4-4h+h^2) \\ 64=a(h^2-4+4h-h^2) \\ 64=a(-4+4h)\rightarrow(5) \end{gathered}`

Divide the equation (4) by (5):

`\begin{gathered} (48)/(64)=(-3+2h)/(-4+4h) \\ \\ (48)/(64)=(3)/(4) \\ \\ 3\cdot(-4+4h)=4\cdot(-3+2h) \\ -12+12h=-12+8h \\ 12h-8h=0 \\ 4h=0 \\ h=0 \end{gathered}`

Substitute into equation 4 to find a:

`\begin{gathered} 48=a\cdot-3 \\ \\ a=(48)/(-3)=-16 \end{gathered}`

Substitute with h and a, into equation 1 to find k

`\begin{gathered} 150=0+k \\ k=150 \end{gathered}`

so, the relation between x and y will be:

`y=-16x^2+150`

Now, we need to find how long is the object in the air?

So, we need to find x when y = 0

`\begin{gathered} y=0 \\ 0=-16x^2+150 \\ 16x^2=150 \\ x^2=(150)/(16)=(75)/(8) \\ \\ x=\sqrt[]{(75)/(8)}\approx3.06186 \end{gathered}`

rounding to the nearest hundredth

So, x = 3.06 seconds

So, the answer will be the object would be in the air for 3.06 seconds