Problem 41 - College Algebra

Question

A. Use the Leading Coefficient Test to determine the graph's end behavior. B. Find the \(x\) -intercepts. State whether the graph crosses the \(x\) -axis, or touches the \(x\) -axis and turns around, at each intercept. C. Find the \(y\) -intercept. 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 maximum number of uning points to check whether it is drawn correctly. $$ f(x)=x^{3}+2 x^{2}-x-2 $$

Step-by-Step Solution

Answer

The end behavior is \(f(x) \rightarrow \infty\) as \(x \rightarrow \infty\) and \(f(x) \rightarrow -\infty\) as \(x \rightarrow -\infty\). The x-intercepts are \(x = -1, 1, -2\) and the graph crosses the x-axis at each of these points. The y-intercept is (0,-2). The graph has neither y-axis symmetry nor origin symmetry. The graph is plotted correctly as the graph has at most 2 turning points and additional points fall correctly on the graph.

Step 1: Determine the end behavior of the polynomial

To determine the end behavior, look at the leading term, which is \(x^{3}\). Since the degree of this polynomial is odd and the leading coefficient is positive, the end behavior is: as \(x \rightarrow -\infty, f(x) \rightarrow -\infty\) and as \(x \rightarrow \infty, f(x) \rightarrow \infty\).

Step 2: Find the x-intercepts of the polynomial

Set \(f(x) = 0\) and solve for \(x\). This gives the equation \(x^{3} + 2x^{2} -x -2 = 0\). Factoring this equation gives \((x +1)(x -1)(x +2) = 0\). Setting each factor equal to zero gives \(x = -1, 1, -2\). The graph crosses the x-axis at these x-intercepts because the multiplicity of each root is 1 (each root is a simple root).

Step 3: Determine the y-intercept of the polynomial

Plug in 0 for \(x\) in \(f(x)\). This gives \(f(0) = -(2)\), so the y-intercept is (0,-2).

Step 4: Determining the symmetry of the graph

A graph has y-axis symmetry if \(f(-x) = f(x)\) for all \(x\) in the domain, origin symmetry if \(f(-x) = -f(x)\) for all \(x\) in the domain, and neither if the graph does not satisfy either tests. Substituting \(x\) with \(-x\) gives \(-x^{3} + 2x^{2} +x-2\), which is not equal to \(x^{3} +2x^{2} -x -2\). Hence, no y-axis symmetry. Also, \(-x^{3} + 2x^{2} +x-2\) is not equal to \(-(x^{3} +2x^{2} -x -2)\), hence no origin symmetry. The graph has neither y-axis nor origin symmetry.

Step 5: Plot additional points and confirm turning points

Let's choose additional points such as \((2, 8)\) - simply plug in the numbers and calculate the function value. Additionally, the degree of \(f(x)\) is 3, so f(x) has at most 2 turning points. A rough sketch of the graph will reveal these turning points, confirming the graph is drawn correctly.

Key Concepts

Leading Coefficient Test

The leading coefficient test helps us understand how the graph of a polynomial behaves at the ends or boundaries. When considering the function, the prominent or leading term is crucial. In our function, this is the term with the highest power:

• Here, the leading term is \(x^3\). The function’s degree, or the highest exponent, is 3, which is odd. The leading coefficient, or the number multiplying the term, is positive (1 in this case).
• Key insight: With an odd-degree and positive leading coefficient, expect one end of the graph to plunge downwards as \(x\) heads left towards negative infinity and the other to climb upwards as \(x\) goes right toward positive infinity.

This pattern is crucial in visualizing polynomial graphs quickly, helping us sketch the general outline of the function's behavior from left to right.

X-Intercepts

The \(x\)-intercepts are the points where the graph crosses or meets the \(x\)-axis. To find them, set the whole polynomial equal to zero:

• The given equation is \(f(x) = x^3 + 2x^2 - x - 2\).
• Factor it into \((x+1)(x-1)(x+2)\) which equals zero.
• This gives the roots \(x = -1, 1, -2\).
• Each of these roots will make the polynomial equal t0 zero, showing us the \(x\)-intercepts.

Due to each factor appearing once (multiplicity of 1), the graph will notably cross the \(x\)-axis at these points rather than merely touching it, which is a key visualization point.

Y-Intercept

The \(y\)-intercept is more straightforward: plug zero into all \(x\) terms of the polynomial and solve. This reveals where the graph meets the \(y\)-axis.

• For this function, substituting \(0\) for \(x\) gives \(f(0) = 0^3 + 2(0)^2 - 0 - 2 = -2\).
• Thus, the \(y\)-intercept is at the point \((0, -2)\), which is where the graph will cross the \(y\)-axis.

This is a precise location on the graph, helping to anchor your sketch visually by providing a clear starting point from which the graph can expand in either direction.

Graph Symmetry

A polynomial graph might show symmetry, useful for simplifying graphing tasks. We test for two types:

• Y-axis Symmetry: Happens when \(f(-x) = f(x)\). It means the graph mirrors itself along the \(y\)-axis.
• This function, when you replace \(x\) with \(-x\), becomes \(-x^3 + 2x^2 + x - 2\), which does not equal \(x^3 + 2x^2 - x - 2\). So, no \(y\)-axis symmetry.
• Origin Symmetry: Occurs if \(f(-x) = -f(x)\). This would indicate the graph can rotate 180° around the origin and look the same.
• Here, \(-x^3 + 2x^2 + x - 2\) is not \(-1\) times the original polynomial, hence, no origin symmetry.

Without either symmetry, the graph does not mirror or flip wholly around the central axis or point. Each side of the graph stands independent, and must be considered on its own.

End Behavior

End behavior helps describe what happens to the graph of the polynomial as \(x\) becomes extremely positive or negative. This aspect allows us to envision how the graph expands or contracts over its domain's limits.

• For \(f(x) = x^3 + 2x^2 - x - 2\), the leading term of \(x^3\) dictates that as \(x\) goes to negative infinity, \(f(x)\) descends, going to negative infinity as well.
• Conversely, as \(x\) heads to positive infinity, \(f(x)\) ascends to positive infinity.

This overarching view of polynomial growth or decline frames how the overall shape and slope transition when moving outwards from center, giving a contextual understanding that helps with naturally orienting the function's entire graph.