Question

the polynomial function $a(t)=0.003462t^{3}-0.03768t^{2}+0.1016t + 0.005$ gives the approximate blood - alcohol concentration in a person $t$ hours after drinking some alcohol on an empty stomach, for $t$ in the interval $0,5$.
a. graph $a(t)$ on $0leq tleq5$.
b. using the graph from part a, estimate the time of maximum alcohol concentration.
c. use the graph from part a to estimate the period in which the concentration of the alcohol in the person exceeds 0.06.
a. which graph below shows $a(t)$? the viewing windows are $0,5$ by $0,0.15$.

Explanation:

Step1: Analyze the polynomial function

The function $A(t)=0.003462t^{3}-0.03768t^{2}+0.1016t + 0.005$ is a cubic polynomial. Cubic polynomials of the form $y = ax^{3}+bx^{2}+cx + d$ with $a>0$ (here $a = 0.003462>0$) generally have a shape that starts from negative - infinity (if considering the whole domain), has a local minimum and a local maximum. For the domain $t\in[0,5]$, we can evaluate the function at some key - points.

Step2: Evaluate the function at endpoints and critical points

First, find the derivative $A'(t)=3\times0.003462t^{2}-2\times0.03768t + 0.1016=0.010386t^{2}-0.07536t + 0.1016$.
To find the critical points, set $A'(t) = 0$. Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $at^{2}+bt + c = 0$, here $a = 0.010386$, $b=-0.07536$, $c = 0.1016$.
$t=\frac{0.07536\pm\sqrt{(-0.07536)^{2}-4\times0.010386\times0.1016}}{2\times0.010386}$.
$t=\frac{0.07536\pm\sqrt{0.005679 - 0.004229}}{0.020772}=\frac{0.07536\pm\sqrt{0.00145}}{0.020772}=\frac{0.07536\pm0.0381}{0.020772}$.
We get two critical points:
$t_1=\frac{0.07536 + 0.0381}{0.020772}=\frac{0.11346}{0.020772}\approx5.46$ (outside the domain $[0,5]$)
$t_2=\frac{0.07536- 0.0381}{0.020772}=\frac{0.03726}{0.020772}\approx1.8$.
Evaluate $A(t)$ at $t = 0$: $A(0)=0.005$.
Evaluate $A(t)$ at $t = 1.8$: $A(1.8)=0.003462\times(1.8)^{3}-0.03768\times(1.8)^{2}+0.1016\times1.8 + 0.005$.
$A(1.8)=0.003462\times5.832-0.03768\times3.24 + 0.18288+0.005$.
$A(1.8)=0.02019 - 0.12208+0.18288 + 0.005=0.086$.
Evaluate $A(t)$ at $t = 5$: $A(5)=0.003462\times(5)^{3}-0.03768\times(5)^{2}+0.1016\times5 + 0.005$.
$A(5)=0.003462\times125-0.03768\times25 + 0.508+0.005$.
$A(5)=0.43275-0.942+0.508 + 0.005=0.00375$.

Step3: Analyze the graph shape

Since $A(0)=0.005$, $A(1.8)\approx0.086$, $A(5)=0.00375$, the function first increases from $t = 0$ to $t\approx1.8$ and then decreases from $t\approx1.8$ to $t = 5$.
a. The graph that shows the function increasing first and then decreasing in the interval $[0,5]$ is the correct one. Without seeing the actual options clearly (but based on the analysis), we know the general shape.
b. The time of maximum alcohol concentration is approximately $t\approx1.8$ hours.
c. To find when $A(t)>0.06$, we set $A(t)=0.003462t^{3}-0.03768t^{2}+0.1016t + 0.005>0.06$.
$0.003462t^{3}-0.03768t^{2}+0.1016t-0.055>0$.
By using a graphing utility or by testing values in the interval $[0,5]$, we find the interval where the graph of $y = 0.003462t^{3}-0.03768t^{2}+0.1016t-0.055$ is above the $t -$axis.
We find that the period is approximately $1.2

Answer:

a. (Without seeing the options, cannot give a specific letter - answer. But it should be a graph that increases then decreases in $[0,5]$)
b. Approximately $1.8$ hours
c. Approximately $1.2