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Find a degree 
Magazine Chapter 3: Problem 266
For the following exercises, graph the potynomial functions. Note \(x\) - and \(y\) -intercepts, multiplicity, and end behavior. $$ h(x)=(x-1)^{3}(x+3)^{2} $$
Short Answer
Expert verified
The graph crosses at \(x=1\) and bounces at \(x=-3\) with y-intercept at \((0, -9)\).
Step by step solution
01
Determine the degree and end behavior
The degree of the polynomial is the sum of the exponents of the factors. Here, we have \((x-1)^3\) and \((x+3)^2\), so the degree is 3 + 2 = 5.For odd-degree polynomials, such as this one, with a positive leading coefficient, the end behavior is as follows:- As \(x \to \infty\), \(h(x) \to \infty\)- As \(x \to -\infty\), \(h(x) \to -\infty\).
02
Identify the x-intercepts and their multiplicities
The x-intercepts occur where each factor is equal to zero:- From \((x-1)^3\), we have an intercept at \(x=1\) with multiplicity 3.- From \((x+3)^2\), we have an intercept at \(x=-3\) with multiplicity 2.
03
Determine the behavior at each x-intercept
- At \(x=1\) (multiplicity 3), the graph touches and crosses the x-axis, creating a point of inflection.- At \(x=-3\) (multiplicity 2), the graph touches the x-axis and bounces off.
04
Find the y-intercept
Set \(x = 0\) in the function to find the y-intercept:\[h(0) = (0-1)^3(0+3)^2 = (-1)^3(3)^2 = -1 \times 9 = -9\]So, the y-intercept is at \((0, -9)\).
05
Sketch the graph
Using the information gathered:- Plot the x-intercepts at \((1, 0)\) and \((-3, 0)\).- Note the y-intercept at \((0, -9)\).- Draw the curve showing the end behavior, crossing at \(x=1\), touching and bouncing at \(x=-3\), and performing according to the polynomial degree (odd) with tails negative on the left and positive on the right.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Degree of Polynomial
The degree of a polynomial is the sum of the exponents on each term. This indicates the highest power of the variable present. In our example, the polynomial \(h(x)=(x-1)^3(x+3)^2\) has two factors: \((x-1)^3\) and \((x+3)^2\).
- The exponent 3 from \((x-1)^3\) and the exponent 2 from \((x+3)^2\) are added together to give us the total degree.
- Thus, the degree is 3 + 2 = 5.
End Behavior
End behavior describes what happens to the values of a polynomial function as \(x\) approaches infinity or negative infinity. This helps us understand how the graph acts at the extremities.For our polynomial, since it's of odd degree with a positive leading coefficient:
- As \(x \to \infty\), \(h(x) \to \infty\)
- As \(x \to -\infty\), \(h(x) \to -\infty\)
X-Intercepts
X-intercepts are points where the graph crosses or touches the x-axis, which means the output \(h(x)\) is equal to zero. To find these points, we solve each factor set to zero.For the polynomial \(h(x)=(x-1)^3(x+3)^2\):
- From \((x-1)^3\), we set \(x-1=0\) which gives \(x=1\). So one x-intercept is at \((1, 0)\).
- From \((x+3)^2\), we set \(x+3=0\) giving \(x=-3\). So another x-intercept is at \((-3, 0)\).
Multiplicity
Multiplicity refers to the number of times a particular x-intercept repeats. In the polynomial function, the multiplicity of a root affects how the graph behaves at that x-intercept.
- A multiplicity of 1 means the graph crosses the x-axis at that intercept.
- A multiplicity greater than 1 indicates the graph touches or bounces off the x-axis.
Y-Intercepts
The y-intercept of a graph is the point where the graph crosses the y-axis. It occurs at \(x=0\), giving us insight into where the function stands when no other particular variables are at play.To find the y-intercept, substitute \(x=0\) into the polynomial:\[h(0) = (0-1)^3(0+3)^2 = (-1)^3(3)^2 = -1 \times 9 = -9\]This calculation shows the y-intercept is at \((0, -9)\).Knowing the y-intercept is vital for understanding the starting point of the graph along the y-axis and for beginning sketches of the polynomial by marking a crucial coordinate point.
