$\blacktriangleright$ Domain and Range First, any function analysis would discuss the domain and range. Perhaps the domain is already stated with the function definition. Otherwise, the function might be described with a formula. Then, the natural or implied domain can be deduced. If the function is defined via a graph, then the implied domain can be deduced visually.

If a formula is provided, then the best method of analysis is to concurrently extract information algebraically and graphically. The algebraic information will help you piece together a sketch. Even a partial sketch will suggest features and characteristics that will direct your algebraic investigation.

If a graph is the only information presented, then there is an unavoidable bubble of inaccuracy to work within. The nature of drawing, the tools used and the person drawing, make inaccuracy simply a part of graphing or decerning information from a graph.

Even with a formula, piecing the graph together may be the only way to obtain clues to certain function aspects.

We want exact values when we can get them. Otherwise, we settle for approximations. But we want algebra first. Then we settle.

$\blacktriangleright$ Formula
If a formula is provided, then it will be built from elementary functions. Begin imaging what the graphs of those pieces look like. You should have all of the important aspects of elementary functions memorized. Begin thinking of important aspects of the function or important graphing points. This will suggest directions the algebraic investigation should take.

$\blacktriangleright$ Zeros
Identify all zeros. This will involve solving equations that you create from your knowledge of the elementary functions. As you identify zeros graph their corresponding intercepts on the graph.

$\blacktriangleright$ Discontinuities and Singularities
Identify discontinuities and singularities. These will show up as breaks in the graph. Describe the function’s behavior around discontinuities and singularities with limit notation.

$\blacktriangleright$ Continuity
The domain, discontinuities, and singularities should begin establishing intervals of continuity. The endpoints or isolation numbers may require additional scrutiny.

$\blacktriangleright$ End-Behavior
Think in terms of very very very big positive and negative numbers. How does the function settle down? Are there limiting values (horizontal asymptotes)? Are there oblique asymptotes? Add these to the graph.

$\blacktriangleright$ Graph
With this information, you should be able to describe a reasonable shape for a graph. It may not have the correct heights. However, it should have the basic shape. It should have important features positioned correctly. It should show ideas of increasing and decreasing. Your ideas should match any graph created with technology.

$\blacktriangleright$ Important Points
Our library of elementary function comes with formulas and graphs. The graphs also list important points that align and position the graph. These points points should be correctly positioned on your graph.

$\blacktriangleright$ Critical Numbers and Extreme Values
Generally speaking, obtaining exact values of local maximums and mimimums and their critical numbers is going to be difficult without Calculus. However, we know a lot about these in special situaitons. Your graph should have strong suggestions of where these extreme values occur in the domain. You can evaluate the function in this area to get an idea of where the graph should be drawn. You also should have a good idea how the elementary functions change.

If you have the derivative, then you can get exact values for critical numbers.

$\blacktriangleright$ Rate-of-Change
With the derivative or with the graph, you can identify intervals where the function increases and decreases. Otherwise, you have your knowledge of the elementary functions and you have a graph.

Complete Anlaysis

Completely analyze $p(t) = (t+3)e^{-2t-3}$.

$p(t)$ is the product of a polynomial and an exponential function. Therefore, its implied domain is all real numbers.

$p(t)$ is in factored form. The exponential factor, $e^{-2t-3}$, does not have a zero. The linear factor has a zero occur when $t+3=0$ or $t=\answer {-3}$.

The exponential factor, $e^{-2t-3}$, only has positive values. Since the exponential factor only has positive values, the sign of $p(t)$ is the same as the sign of $t+3$. Therefore, $p(t)$ is positive on $\left ( \answer {-3}, \infty \right )$ and negative on $\left ( -\infty , \answer {-3} \right )$.

As for end-behavior, this exponential factor dominates over the polynomial factor. The exponential values tend to $0$ when the exponent takes on large negative values. That would be when $t$ tends to $\infty$. The exponential values tend to $\infty$ when the exponent takes on large positive values. That would be when $t$ tends to $-\infty$.

This tells us that

• $p(t)$ tends to $0$ from the positive side as $t \to \infty$.
• $p(t)$ tends to $-\infty$ as $t \to -\infty$.
• The graph crosses the horizontal axis at $\left ( \answer {-3}, \answer {0} \right )$.

Neither polynomials nor exponentials have discontinuities or singularities. Therfore, $p(t)$ is continuous on $(-\infty , \infty )$.

We can begin to piece together a sketch of the graph.

The graph suggests that there is a hill between $-4$ and $0$, which means there is a global maximum somewhere on $(-4, 0)$. Without a derivative, we cannot pin this critical number down. But we now know enough to make a nice sketch of the function.

There is no vertical asymptote. Exponential functions do not have singularities and neither do polynomials. The domain is $(-\infty , \infty )$ and the graph continues down and to the left.

$y = 0$ is a horizontal asymptote.

From the graph we can estimate the critical number as approximately $-2.5$.

• $p(t)$ increases on $(-\infty , -2.5]$.
• $p(t)$ decreases on $[-2.5, \infty )$.

$p(t)$ has a global maximum of approximately $3.7$ at $-2.5$.

$\lim \limits _{t \to -\infty } p(t) = -\infty$

$\lim \limits _{t \to \infty } p(t) = 0$

If we are given the derivative, we could have determined the critical number and maximum value.

$p'(t) = -(2t+5) e^{-2t-3}$, which equals $0$ at $-2.5$. Our estimation was right on target.

The maximum value is $p(-2.5) = (-2.5+3)e^{-2(-2.5)-3} = 0.5 \, e^{2}$. This is approximately $3.6945$.