Problem 36 - Precalculus : Building Concepts and Connections

Question

For each polynomial function, find (a) the end behavior; (b) the \(y\) -intercept; (c) the \(x\) -intercept(s) of the graph of the function and the multiplicities of the real zeros; (d) the symmetries of the graph of the function, if any; and (e) the intervals on which the function is positive or negative. Use this information to sketch a graph of the function. Factor first if the expression is not in factored form. $$f(x)=-x^{3}-2 x^{2}-x$$

Step-by-Step Solution

Answer

The function is \(f(x) = -x(x+1)^2\), its end behavior is up to the left and down to the right. The y-intercept is at (0,0), x-intercepts are at x = 0 and x = -1 with multiplicities 1 and 2, respectively. There's no symmetry about the y-axis or the origin. \(f(x) > 0\) on the interval \((-∞, 0]\) and \(f(x) < 0\) on the interval \([0, +∞)\).

Step 1: Factorize the polynomial

The polynomial can be factorized by factoring out the greatest common factor, which is \(x\). So, we get: \(f(x) = -x(x^2+2x+1)\). Now, further factorize the quadratic expression to get \(f(x) = -x(x+1)^2\).

Step 2: Find end behavior

To find the end behavior, examine the leading term, which is \(-x^3\). As the degree of the polynomial is odd and the coefficient of the leading term is negative, as \(x\) approaches positive infinity (\(x \to +\infty\)), \(f(x)\) approaches negative infinity (\(f(x) \to -\infty\)). Similarly, as \(x\) approaches negative infinity (\(x \to -\infty\)), \(f(x)\) approaches positive infinity (\(f(x) \to +\infty\)). So the end behavior of the function is 'upwards to the left and downwards to the right'.

Step 3: Find the y-intercept

The y-intercept is the point where the curve crosses the y-axis. For any function, this occurs when \(x = 0\). So, substitute \(x = 0\) into the equation we get \(f(0) = -0(0 + 1)^2 = 0\). Therefore, the y-intercept is at (0,0).

Step 4: Find the x-intercept(s)

The x-intercept is the point where the curve crosses the x-axis. This happens when \(f(x) = 0\). So, from the equation \(f(x) = -x(x+1)^2 = 0\), we can find the x-intercepts are at \(x = 0\) and \(x = -1\) with multiplicities 1 and 2 respectively.

Step 5: Determine the symmetry

The function is symmetrical if it is either even or odd. A function is said to be even if \(f(-x) = f(x)\), and it’s said to be odd if \(f(-x) = -f(x)\). Computing \(f(-x)\) for the given function gives \(-(-x)[(-x)+1]^2\), which simplifies to \(-x(x-1)^2\). This doesn't equal both \(f(x)\) and \(-f(x)\), so the graph of the function is neither symmetric about the y-axis nor the origin.

Step 6: Find where the function is positive and negative

Sketching a sign chart would provide regions where the function is positive and negative. The critical points are -1 and 0 (x-intercepts), test points we can select are -2 , -0.5, and 1. For \(x <-1\), i.e \(x = -2\), \(f(x)>0\), for \(-1< x <0\), i.e \(x = -0.5\), \(f(x)>0\), and for \(x >0\), i.e \(x = 1\), \(f(x)<0\). Thus \(f(x) > 0\) on the interval \((-∞, 0]\) and \(f(x) < 0\) on the interval \([0, +∞)\).

Step 7: Sketch the graph

First, plot the points for the intercepts on a graph. Then draw the curve according to the end behaviors, intercepts, symmetry, and the regions where \(f(x)>0\) and \(f(x)<0\). The graph will start from quadrant II, move downwards to touch the x-axis at x=-1, bounce back above the x-axis, cross the x-axis at x=0 downward, and continue to negative infinity.

Key Concepts

End Behavior

Understanding the end behavior of a polynomial function involves observing what happens to the function's values as the input, \(x\), becomes very large or very small. For the polynomial \(f(x) = -x^3 - 2x^2 - x\), the leading term is \(-x^3\). Since it has an odd degree and a negative leading coefficient, the graph will behave as follows:

• As \(x\) approaches positive infinity \((x \to +\infty)\), \(f(x)\) approaches negative infinity \((f(x) \to -\infty)\).
• Conversely, as \(x\) approaches negative infinity \((x \to -\infty)\), \(f(x)\) approaches positive infinity \((f(x) \to +\infty)\).

This tells us that the graph starts high and ends low, moving **upwards to the left and downwards to the right**.

Y-Intercept

The y-intercept of a function is found where the graph crosses the y-axis. This point is identified by setting \(x = 0\) in the function. For \(f(x) = -x(x+1)^2\), substituting \(x = 0\) gives
\[f(0) = -0(0 + 1)^2 = 0\]
Hence, the y-intercept is at the origin, **(0, 0)**.

X-Intercept

To find the x-intercepts, we examine where \(f(x) = 0\). Here, we solve the equation
\[-x(x+1)^2 = 0\]
which gives solutions at \(x = 0\) and \(x = -1\). These are the points where the graph crosses the x-axis.

The intercepts have multiplicities, which indicate how the graph behaves at these points:

• \(x = 0\) has a multiplicity of **1**, meaning the graph crosses the x-axis at this point.
• \(x = -1\) has a multiplicity of **2**, indicating that the graph touches the x-axis and bounces back.

Symmetry in Graphs

For symmetry, we examine if the function is even or odd. A function is even if \(f(-x) = f(x)\) and odd if \(f(-x) = -f(x)\). Substituting \(-x\) into our function gives:
\[-(-x)((-x)+1)^2 = -x(x-1)^2\]
This result is neither equal to \(f(x)\) nor \(-f(x)\), indicating that the function lacks symmetry about the y-axis or the origin.

Function Positivity and Negativity

To determine where the function is positive or negative, we test intervals between critical points, especially around the x-intercepts at \(-1\) and \(0\).

• For \(x < -1\), such as \(x = -2\), \(f(x) > 0\).
• For \(-1 < x < 0\), such as \(x = -0.5\), \(f(x) > 0\).
• For \(x > 0\), such as \(x = 1\), \(f(x) < 0\).

Therefore, the function is positive in the intervals \((-\infty, 0]\) and negative on \([0, +\infty)\).