Respuesta :

Solution:

Given the graph below:

3) Horizontal asymptote

The above graph is an exponential function expressed as

[tex]P(t)=a(b)^t\text{ ---- equation 1}[/tex]

To evaluate the horizontal asymptote, we need to evaluate the values of a and b.

Thus,

[tex]\begin{gathered} From\text{ the graph, when t=0, P\lparen t\rparen =5} \\ Thus,\text{ substitute these valuaes into equation 1} \\ 5=a(b)^0 \\ 5=a\times1 \\ \Rightarrow a=5 \\ \end{gathered}[/tex]

Substitute the value of a into equation 1.

Thus, we have

[tex]P(t)=5(b)^t----\text{ equation 2}[/tex]

Also,

[tex]\begin{gathered} when\text{ t=1, P\lparen t\rparen=10} \\ substitute\text{ these values into equation 2} \\ 10=5(b)^1 \\ \Rightarrow10=5b \\ divide\text{ both sides by the coefficient of b, which is 5.} \\ \frac{10}{5}=\frac{5b}{5} \\ \Rightarrow b=2 \end{gathered}[/tex]

Substitute the value of b into equation 2.

Thus, the exponential function P(t) becomes

[tex]P(t)=5(2)^t----\text{ equation 3}[/tex]

To evaluate the horizontal asymptote, we take the limit of the function P(t) as t tends to infinity (either positive or negative).

Thus, we have

[tex]\lim_{t\to-\infty}P(t)=\lim_{t\to-\infty}(5(2)^t)=0[/tex]

Hence, the horizontal asymptote of the function is at

[tex]P(t)=0[/tex]

4) Domain of the function:

The domain of the P(t) function is the set of input values for which the P(t) function is real and defined.

Thus, the domain of the function is

[tex]-\infty\:

5) Range of the function:

The range of the function P(t) is the set of dependable values for which the function P(t) is defined.

Thus, the range of the function is

[tex]P(t)>0[/tex]

6) y-intercept.

The y-intercept is evaluated to be the value of P(t) when t equals zero.

Thus, at the y-intercept,

[tex]x=0[/tex]

Thus, from the exponential function P(t),

[tex]\begin{gathered} P(t)=5(2)^t \\ at\text{ the y-intercept, x=0} \\ P(t)=5(2)^0=5\times1 \\ \Rightarrow P(t)=5 \end{gathered}[/tex]

Hence, the y-intercept of the function is 5.

This implies that at the snake population was 5 at the year the Scientists started keeping tracks of the population.

7) Point on the graph when x equals 1.

From the graph, the point (x,y) is

[tex](1,10)[/tex]

This implies that at the end of the first year the snake population is 10.

8) The function is an exponential function expressed as

[tex]P(t)=5\cdot2^t[/tex]

9) To fill the table shown below:

we substiyute the respective values of t into the function P(t)

Ver imagen HershelF562428
Ver imagen HershelF562428
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