Problem 35 - College Algebra

Question

In Exercises \(35-50\) a. Use the Leading Coefficient Test to determine the graphs end behavior. b. Find \(x\) -intercepts by setting \(f(x)=0\) and solving the resulting polynomial equation. State whether the graph crosses the \(x\)-axis, or touches the \(x\)-axis and turns around, at each intercept. c. Find the \(y\) -intercept by setting \(x\) equal to 0 and computing \(f(0)\) d. Determine whether the graph has \(y\) -axis symmetry, origin symmetry, or neither. e. If necessary, find a few additional points and graph the function. Use the fact that the maximum number of turning points of the graph is \(n-1\) to check whether it is drawn correctly. $$f(x)=x^{3}+2 x^{2}-x-2$$

Step-by-Step Solution

Answer

The end behavior of the function goes down on the left and up on the right. There are x-intercepts where the function crosses the x-axis, which can be found by solving the equation and the y-intercept is at -2. The function has no symmetry. Additional points can be found to confirm the graph, which should have a maximum of 2 turning points as per the degree of the polynomial.

Step 1: Determine the end behavior

Using the Leading Coefficient Test, since the power of the leading term is odd and its coefficient is positive, therefore the end behavior of the function is: down on the left (\(-\infty\)) and up on the right (\(+\infty\)).

Step 2: Find the x-intercepts

To find the x-intercepts, set \(f(x) = 0\). Thus, \(x^{3} + 2x^{2} - x - 2 = 0\). Solving this equation gives the roots or x-intercepts. The function crosses thr x-axis at each of these points.

Step 3: Find the y-intercept

The y-intercept is found by setting \(x = 0\) in \(f(x)\). Substituting we get \(f(0) = -2\). So, the y-intercept is at -2.

Step 4: Determine symmetry

By substituting \(-x\) for \(x\) in \(f(x)\) and simplifying, if the result is equal to \(f(x)\) then it has y-axis symmetry. If the result equals to \(-f(x)\) then it has origin symmetry. If neither, then the graph has no symmetry. The given function is asymmetric around both the y-axis and the origin.

Step 5: Find additional points and graph

Choose some values for \(x\) and calculate corresponding \(f(x)\) to get additional points. Sketch the graph using these points, the behaviour from step 1 and intercepts from steps 2 and 3. For a third degree polynomial, the maximum number of turning points is \(3 - 1 = 2\). Confirm that the graph is correct by this rule.

Key Concepts

Leading Coefficient Test

The Leading Coefficient Test is an essential tool when working with polynomial functions. This test helps to predict how the graph of a polynomial behaves as you move towards the ends of the graph, either directionally positive (to the right) or negative (to the left).

For any polynomial, the leading term is the one with the highest power of the variable. Here, in the function \(f(x) = x^3 + 2x^2 - x - 2\), the leading term is \(x^3\), where the exponent is 3 (odd), and the leading coefficient is 1 (positive).

According to the Leading Coefficient Test:

• When the leading term has an odd power and a positive coefficient, the graph starts down to the left \(-\infty\) and rises up to the right \(+\infty\).

Thus, knowing these characteristics helps you sketch the broader shape of the polynomial's graph without needing to plot numerous points.

x-intercepts

Finding the \(x\)-intercepts of a polynomial function is crucial because it provides points where the graph intersects the \(x\)-axis. These are also the roots or zeros of the function.

To find them, you set the function \(f(x)\) equal to zero and solve for \(x\). For our function, this means solving the equation \(x^3 + 2x^2 - x - 2 = 0\). Once you have the solutions:

• Each solution is an \(x\)-intercept.
• The graph crosses the \(x\)-axis at each intercept because the polynomial changes signs (crosses) as it passes through these points.

Understanding where these crossovers happen can help you confirm the function's graph is correct as it smoothly transitions through these intercepts.

y-intercepts

A \(y\)-intercept of a polynomial function is where the graph crosses the \(y\)-axis. It is determined by evaluating the function when \(x\) is zero.

In our exercise, if you substitute \(x = 0\) in \(f(x) = x^3 + 2x^2 - x - 2\), you get:
\[ f(0) = (0)^3 + 2(0)^2 - (0) - 2 = -2 \]
This result tells us that the \(y\)-intercept is at the point \((0, -2)\). Hence, the graph goes through \(-2\) on the \(y\)-axis. This intercept provides an additional fixed point, essential when sketching the graph, as it locks down one more location the graph must pass through.

Graph Symmetry

Symmetry in the graph of a polynomial can simplify its analysis and sketching. We typically check for two types of symmetry: - **Y-axis Symmetry:** This means the graph reflects identically across the \(y\)-axis. We test this by substituting \(-x\) into \(f(x)\) and see if we get back the original function (\(f(-x) = f(x)\)).- **Origin Symmetry:** This refers to the graph reflecting through the origin. It applies if \(f(-x) = -f(x)\) holds true.

For the given function, \(f(x) = x^3 + 2x^2 - x - 2\), substituting \(-x\) does not result in either \(f(x)\) or \(-f(x)\).

• This indicates no symmetry around the \(y\)-axis or origin.

Not having symmetry simplifies our graph sketching process, removing the need to look for these reflective properties.