Problem 34 - 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. $$f(x)=-3(x-2)^{2}(x+1)^{2}$$

Step-by-Step Solution

Answer

The end behavior of \(f(x)=-3(x-2)^{2}(x+1)^{2}\) is \( (\infty, -\infty)\), the function intercepts the y-axis at the point (0, -12), and the x-axis at x=-1 and x=2. The multiplicities of -1 and 2 are both 2. There are no symmetries, and the function is negative on the intervals (-∞, -1) and (2, ∞), and positive on (-1, 2). The graph can be sketched based on this information.

Step 1: Find the End Behavior

Observe the leading term of \(f(x)\). In our case, the leading term would be \(-3x^4\), since the highest power of x is 4 and the coefficient is -3. As the highest power of the x term is even, the end behavior of the function will either be up for both ends or down for both ends. The coefficient of this term is negative, which means the end behavior will be up on the left and down on the right--written as \( (\infty, -\infty)\).

Step 2: Find the y-intercept

The y-intercept is found by setting x to zero in the function. So, \(f(0)=-3(0-2)^2(0+1)^2 = -3(-2)^2(1)^2 = -12\). Therefore, the y-intercept is at point (0,-12).

Step 3: Find the x-intercepts and multiplicities

The x-intercept(s) of the function are the real zeros of the function, which can be found by setting the function to zero and solving for x. From the factored form of our function, it can be seen that the solutions are \(x=2\) and \(x=-1\). The exponent on each factor corresponds to the multiplicity of the solution, so they each have a multiplicity of 2 in this case.

Step 4: Find the symmetries

For even-powered polynomial functions, if every term in the function has an even power or the function is constant, the function is symmetric about the y-axis. In this case, there are no symmetries in the graph, since there are odd powered terms present.

Step 5: Finding the interval of positivity and negativity

The polynomial is negative for the interval \((-∞, -1)\) since the sign of each factor changes at each zero. Beyond the zero at x=-1, there are an even number of negative factors leading to a positive value, so the function is positive on the interval \((-1, 2)\). Beyond the zero at x=2, there are again an odd number of negative factors leading to a negative value, so the function is again negative on the interval \((2, ∞)\).

Step 6: Sketch the Graph

The graph should start from the left upper quadrant due to the negative leading term. It passes through the y-intercept at (0,-12) and bends at the x-intercepts without crossing the x-axis (since the zeros have even multiplicity). The graph will touch the x-axis at the zeros x=-1 and x=2, and the end behavior shows it ending in the lower right quadrant.

Key Concepts

Understanding End Behavior in Polynomial Functions

When we talk about the end behavior of polynomials, we're referring to what parts of the graph look like as the x-values move towards positive or negative infinity. Let's break this down using the example polynomial \( f(x) = -3(x - 2)^2(x + 1)^2 \). The most crucial part to consider is the leading term, which governs the end behavior. Here, it is \(-3x^4\), taken from multiplying the highest power of x from each factor.

• The power of the leading term, 4, is even. Normally, for polynomials with even powers, the graph extends in the same direction (either up or down) as x moves to both positive and negative infinity.
• Since the coefficient of \(x^4\) is negative (-3), the branches of the graph will both head downward. As a result, the graph behaves by going upwards as we move to the left towards negative infinity, and then it will descend downwards on the right, towards positive infinity.

This results in an end behavior described as \((\infty, -\infty)\). Monitoring the symmetry and power of the leading term can quickly help one determine this core characteristic of polynomial functions.

Finding the y-intercept of a Polynomial Function

The y-intercept of a function is where the graph crosses the y-axis. Mathematically, this is calculated by setting \(x = 0\) in the polynomial function and solving for \(y\). For the given function \( f(x) = -3(x-2)^2(x+1)^2 \), you substitute zero for every \(x\):

• The expression simplifies to: \( f(0) = -3(0-2)^2(0+1)^2 = -3(-2)^2(1)^2 \).
• This evaluates to \(-3 \times 4 \times 1 = -12\).

Thus, the y-intercept of this polynomial is at the point \((0, -12)\). The point gives us a crucial location on the graph and serves as the anchor from which other features of the graph can be laid out.

Understanding x-intercepts and Multiplicities of Polynomial Functions

The x-intercepts of a polynomial function are points where the graph crosses or touches the x-axis. For the function \( f(x) = -3(x - 2)^2(x + 1)^2 \), these are obtained by finding the values of \(x\) that make the function equal to zero.

• The roots of the function are \(x = 2\) and \(x = -1\).
• Each of these intercepts has a multiplicity of 2, derived from the exponents in the factored form.

The multiplicity tells us how the graph behaves at each intercept:

• If the multiplicity is even, like in our example, the graph "touches" the axis and turns back without crossing the x-axis at these points.
• If the multiplicity were odd, the graph would "cross" the x-axis at these intercepts.

Understanding x-intercepts and their multiplicities aids in identifying crucial points of contact on the graph and provides valuable insights into the shape and direction of the polynomial's graph.