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Arlecino [84]
3 years ago
12

Tom, Larry, Roberta, and John enjoy roller Coasters. Whenever a new roller Coaster opens near their town,

Mathematics
1 answer:
vazorg [7]3 years ago
5 0

Answer:

  • <u>a) degree 3</u>
  • <u>b) 3 times</u>
  • <u>c) cubic</u>

<u></u>

Explanation:

The function that represents the first drop of the roller coaster is:

       h(t)=t^3-12t^2+35t,

where t is the  time in seconds and h is the height in feet.

<u></u>

<u>a) What is the degree of the function?</u>

The degree of a polynomial is the highest power of the variable, t. Thus, the degree of the function is 3.

<u>b) According to the degree of the function, how many times should the roller coaster hit the ground?</u>

The number of complex roots of a polynomial is equal to the degree of the polynomial. Thus, in this case this polynomial has 3 roots. Each root represents the value of the polynomial that makes it equal to zero, this is:

     t^3-12t^2+35t=0

Thus, since the polynomial function represents the height of the, the value zero represents when the roller coaster hits the ground.

To learn if the three roots are real numbers you can solve the equation, which may be done by factoring the polynomial.

Take t out as common factor:

  • t(t² - 12t + 35)

Find two numbers that add up -12 and its product is 35. Those are -7 and -5. Hence, use the factors t - 12 and t - 5:

  • t ( t - 12) (t - 5) = 0  This is the factored form of the polynomial

Whose solutions are t =0, t = 12, and t = 5. Three real numbers.

Hence, the roller coaster hits the ground three times: first at start, then  at 5 seconds, and finally at 12 seconds.

<u>c) Will the roller coaster be in the form of a linear, quadratic, cubic or quartic graph?</u>

It will be in the form of a cubic graph, becasue a linear graph corresponds to a polynomial of degree 1, a quadratic graph corresponds to a polynomial of degree 2, a cubic graph corresponds to a polynomial of degree 3, and a quartic graph corresponds to a polynomial of degree 4.

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(a)
The inverse is when you swap the variables and solve for y.
g(t) = 2t - 1 (Note: g(t) represents y)
rewrite as: y = 2t - 1
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(b)
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h(t) = 4t + 3
rewrite as: y = 4t + 3
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(c)
g^{-1} ( h^{-1}(t)) =  g^{-1} (\frac{t - 3}{4})
replace all t's in the g^{-1}(t) equation with \frac{t - 3}{4}
 g^{-1} (\frac{t - 3}{4}) = \frac{ \frac{t-3}{4} + 1}{2}
= \frac{ \frac{t-3}{4} +  \frac{4}{4}}{2} = \frac{ \frac{t - 3 + 4}{4}}{2} = \frac{ \frac{t + 1}{4}}{2} =  \frac{t + 1}{8}
Answer for (c): g^{-1} ( h^{-1}(t)) = \frac{t + 1}{8}

 (d)
h(g(t)) = h(2t - 1) = 4(2t - 1) + 3 = 8t - 4 + 3 = 8t - 1
Answer for (d): h(g(t)) = 8t - 1

(e)
h(g(t)) = 8t - 1
   y = 8 t - 1
   t = 8y - 1
  t + 1 = 8y
\frac{t + 1}{8} = y
Answer for (e): inverse of h(g(t)) = \frac{t + 1}{8}
 
























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