Graphical Behavior at Intercepts

If we graph the function f(x) = (x + 3)(x -2)²(x + 1)³, notice that the behavior at each of the horizontal intercepts is different. At the horizontal intercept x = –3, coming from the (x + 3) factor of the polynomial, the graph passes directly through the horizontal intercept. The factor is linear (has a power of 1), so the behavior near the intercept is like that of a line—it passes directly through the intercept. We call this a single zero, since the zero corresponds to a single factor of the function. At the horizontal intercept x = 2, coming from the (x – 2)² factor of the polynomial, the graph touches the axis at the intercept and changes direction. The factor is quadratic (degree 2), so the behavior near the intercept is like that of a quadratic—it bounces off of the horizontal axis at the intercept. Since (x – 2)² = (x – 2)(x – 2), the factor is repeated twice, so we call this a double zero. At the horizontal intercept x = –1, coming from the (x + 1)³ factor of the polynomial, the graph passes through the axis at the intercept, but flattens out a bit first. This factor is cubic (degree 3), so the behavior near the intercept is like that of a cubic, with the same "S" type shape near the intercept that the toolkit x³ has. We call this a triple zero. By utilizing these behaviors, we can sketch a reasonable graph of a factored polynomial function without needing technology.

Graphical Behavior of Polynomials at Horizontal Intercepts

If a polynomial contains a factor of the form (x – h)^p, the behavior near the horizontal intercept h is determined by the power on the factor. For higher even powers 4,6,8, etc., the graph will still bounce off of the horizontal axis but the graph will appear flatter with each increasing even power as it approaches and leaves the axis. For higher odd powers, 5,7,9, etc., the graph will still pass through the horizontal axis but the graph will appear flatter with each increasing odd power as it approaches and leaves the axis.

Example 1

Sketch a graph of f(x) = –2(x + 3)²(x – 5). This graph has two horizontal intercepts. At x = –3, the factor is squared, indicating the graph will bounce at this horizontal intercept. At x = 5, the factor is not squared, indicating the graph will pass through the axis at this intercept. Additionally, we can see the leading term, if this polynomial were multiplied out, would be –2x³, so the long-run behavior is that of a vertically reflected cubic, with the outputs decreasing as the inputs get large positive, and the inputs increasing as the inputs get large negative. To sketch this we consider the following: As x → –∞ the function f(x) → ∞ so we know the graph starts in the 2nd quadrant and is decreasing toward the horizontal axis. At (–3, 0) the graph bounces off of the horizontal axis and so the function must start increasing. At (0, 90) the graph crosses the vertical axis at the vertical intercept. Somewhere after this point, the graph must turn back down or start decreasing toward the horizontal axis since the graph passes through the next intercept at (5,0). As x → ∞ the function f(x) → –∞ so we know the graph continues to decrease and we can stop drawing the graph in the 4th quadrant. Using technology we can verify that the resulting graph will look like:

Solving Polynomial Inequalities

One application of our ability to find intercepts and sketch a graph of polynomials is the ability to solve polynomial inequalities. It is a very common question to ask when a function will be positive and negative. We can solve polynomial inequalities by either utilizing the graph, or by using test values.

Example 2

Solve (x + 3)(x + 1)²(x – 4) > 0 As with all inequalities, we start by solving the equality (x + 3)(x + 1)²(x – 4) = 0, which has solutions at x = –3, –1, and 4. We know the function can only change from positive to negative at these values, so these divide the inputs into 4 intervals. We could choose a test value in each interval and evaluate the function at each test value to determine if the function is positive or negative in that interval

Interval	Test x in interval	f(test value)	>0 or <0?
x < –3	–4	72	> 0
–3 < x < –1	–2	–6	< 0
–1 < x < 4	0	–12	< 0
x > 4	5	288	> 0

On a number line this would look like: From our test values, we can determine this function is positive when x < –3 or x > 4, or in interval notation [latex]\left(-\infty,-3\right)\cup\left(4,\infty\right)[/latex] We could have also determined on which intervals the function was positive by sketching a graph of the function. We illustrate that technique in the next example

Example 3

Find the domain of the function [latex]\displaystyle{v}{({t})}=\sqrt{{{6}-{5}{t}-{t}^{{2}}}}[/latex] . A square root is only defined when the quantity we are taking the square root of, the quantity inside the square root, is zero or greater. Thus, the domain of this function will be when 6 – 5 t – t² $latex \ge$ 0. Again we start by solving the equality 6 – 5t – t² = 0. While we could use the quadratic formula, this equation factors nicely to (6 + t)(1 – t) = 0, giving horizontal intercepts t = 1 and t = –6. Sketching a graph of this quadratic will allow us to determine when it is positive. From the graph we can see this function is positive for inputs between the intercepts. So [latex]6-5t-{{t}^{2}}\ge0[/latex] for [latex]-6\let\le1[/latex], and this will be the domain of the v(t) function.

Try it Now 1

Given the function use the methods that we have learned so far to find the vertical and horizontal intercepts, determine where the function is negative and positive, describe the long run behavior and sketch the graph without technology.

Writing Equations using Intercepts

Since a polynomial function written in factored form will have a horizontal intercept where each factor is equal to zero, we can form a function that will pass through a set of horizontal intercepts by introducing a corresponding set of factors. Factored Form of Polynomials If a polynomial has horizontal intercepts at x = x₁,x₂,...,x_n, then the polynomial can be written in the factored form f(x) = a(x – x₁)^pn(x – x₂)^pn...(x – x_n)^pn where the powers p_i on each factor can be determined by the behavior of the graph at the corresponding intercept, and the stretch factor a can be determined given a value of the function other than the horizontal intercept.

Example 4

Write a formula for the polynomial function graphed here. This graph has three horizontal intercepts: x = –3, 2, and 5. At x = –3 and 5 the graph passes through the axis, suggesting the corresponding factors of the polynomial will be linear. At x = 2 the graph bounces at the intercept, suggesting the corresponding factor of the polynomial will be 2nd degree (quadratic). Together, this gives us f(x) = a(x + 3)(x – 2)²(x – 5) To determine the stretch factor, we can utilize another point on the graph. Here, the vertical intercept appears to be (0,–2), so we can plug in those values to solve for a: [latex-display]\displaystyle-{2}={a}{({0}+{3})}{({0}-{2})}^{{2}}{({0}-{5})}[/latex-display] [latex-display]\displaystyle-{2}=-{60}{a}[/latex-display] [latex-display]\displaystyle{a}=\frac{{1}}{{30}}[/latex-display] The graphed polynomial appears to represent the function [latex]\displaystyle{f{{({x})}}}=\frac{{1}}{{30}}{({x}+{3})}{({x}-{2})}^{{2}}{({x}-{5})}[/latex].

Try it Now 2

Given the graph, write a formula for the function shown.

Estimating Extrema

With quadratics, we were able to algebraically find the maximum or minimum value of the function by finding the vertex. For general polynomials, finding these turning points is not possible without more advanced techniques from calculus. Even then, finding where extrema occur can still be algebraically challenging. For now, we will estimate the locations of turning points using technology to generate a graph.

Example 5

An open-top box is to be constructed by cutting out squares from each corner of a 14cm by 20cm sheet of plastic then folding up the sides. Find the size of squares that should be cut out to maximize the volume enclosed by the box. We will start this problem by drawing a picture, labeling the width of the cut-out squares with a variable, w. Notice that after a square is cut out from each end, it leaves a (14 – 2w) cm by (20 – 2w) cm rectangle for the base of the box, and the box will be w cm tall. This gives the volume: V(w) = (14 – 2w)(20 – 2w)w = 280w – 68w² + 4w³ Using technology to sketch a graph allows us to estimate the maximum value for the volume, restricted to reasonable values for w: values from 0 to 7. From this graph, we can estimate the maximum value is around 340, and occurs when the squares are about 2.75cm square. To improve this estimate, we could use advanced features of our technology, if available, or simply change our window to zoom in on our graph. From this zoomed-in view, we can refine our estimate for the max volume to about 339, when the squares are 2.7cm square.

Try it Now 3

Use technology to find the maximum and minimum values on the interval [–1, 4] of the function [latex]\displaystyle{f{{({x})}}}=-{0.2}{({x}-{2})}^{{3}}{({x}+{1})}^{{2}}{({x}-{4})}[/latex].

Important Topics of this Section

• Short Run Behavior
  • Intercepts (Horizontal & Vertical)
• Methods to find Horizontal intercepts
  • Factoring Methods
  • Factored Forms
  • Technology
• Graphical Behavior at intercepts
• Single, Double and Triple zeros (or power 1, 2, and 3 behaviors)
• Solving polynomial inequalities using test values & graphing techniques
• Writing equations using intercepts
• Estimating extrema

Try it Now Answers

1. The function is negative on (–∞, –2) and (0, 3)The function is positive on (–2, 0) and (3,∞)The leading term is x³ so as x → –∞, g(x) → –∞ and as x → ∞, g(x) → ∞
2. [latex]\displaystyle{f{{({x})}}}=-\frac{{1}}{{8}}{({x}-{2})}^{{3}}{({x}+{1})}^{{2}}{({x}-{4})}[/latex]
3. The minimum occurs at approximately the point (0, –6.5), and the maximum occurs at approximately the point (3.5, 7).

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David Lippman and Melonie Rasmussen, Open Text Bookstore, Precalculus: An Investigation of Functions, "Chapter 3: Polynomial and Rational Functions," licensed under a CC BY-SA 3.0 license.