Problem 27
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. $$f(x)=x^{2}(x-1)$$
Step-by-Step Solution
Verified Answer
End behavior: \(f(x) \rightarrow +\infty\) as \(x \rightarrow \infty\) and \(f(x) \rightarrow -\infty\) as \(x \rightarrow -\infty\). Y-intercept: (0,0). X-intercepts: (0,0), (1,0) with corresponding multiplicities 2, 1. No symmetry. Intervals where function is positive: (0,1); negative: (-\infty, 0) and (1, \infty).
1Step 1: Determine the End Behavior
The end behavior of a function is determined by the degree and the leading coefficient of the polynomial. Here, the degree is 3 (sum of the powers of \(x\)) and leading coefficient is 1. Since the degree is odd and the leading coefficient is positive, as \(x\) approaches infinity (\(x \rightarrow \infty\)), \(f(x)\) goes to +infinity (\(f(x) \rightarrow +\infty\)) and as \(x\) approaches negative infinity (\(x \rightarrow -\infty\)), \(f(x)\) goes to -infinity (\(f(x) \rightarrow -\infty\)).
2Step 2: Find the y-intercept
The y-intercept of the function is the point where the graph crosses the y-axis. This occurs when \(x = 0\). Substituting \(x = 0\) in \(f(x) = x^{2}(x-1)\), we get \(f(0) = 0*0(0-1) = 0\). So, the y-intercept is (0, 0).
3Step 3: Find the x-intercepts and multiplicities
The x-intercepts of the function are the points where the graph crosses the x-axis. These points correspond to the real zeros of the function. The zeros are the values of \(x\) for which \(f(x) = 0\). Setting \(f(x) = x^{2}(x-1) = 0\), we get \(x = 0\) and \(x = 1\) as the roots. The multiplicity of the root 0 is 2 (since \(x^{2}\)), and the multiplicity of the root 1 is 1 (since \(x-1\)). Thus, the x-intercepts are (0, 0) and (1, 0).
4Step 4: Calculate Symmetry of the Graph
If a function is even, then it will have y-axis symmetry (i.e., \(f(x) = f(-x)\)). If a function is odd, it will have origin symmetry (i.e., \(f(x) = -f(-x)\)). With \(f(x) = x^{2}(x-1)\), neither condition holds true. Thus, the graph has no symmetry.
5Step 5: Finding the intervals where the function is positive or negative
To find out the intervals where the function is positive or negative, we glean information from our roots and leading coefficient. For \(x<0\), function will be negative (from end behavior), between 0 and 1 it will be positive (since the sign changes at each root, and multiplicity of 0 is even), and for \(x>1\) it will be negative.
6Step 6: Sketching the function
Based on your information from steps 1-5, graph the function. As \(x\) gets large in the positive or negative direction, the end behavior identified in step 1 is confirmed. The y-intercept determined in step 2 and the x-intercepts found in step 3 are marked on the x and y axes. Lastly, the function's positivity or negativity based on intervals determined in step 5 are marked on the graph.
Key Concepts
End Behavior of Polynomial FunctionsX-Intercepts and Zeros of PolynomialsY-Intercept of GraphIntervals of Positivity and Negativity
End Behavior of Polynomial Functions
When it comes to understanding the graph of a polynomial function, recognizing the end behavior is fundamental. Think of end behavior as the function's ultimate journey as the input value, which is the value of the variable typically represented as 'x', moves towards extreme values on both sides of the number line. It tells us what happens to the values of the function, often denoted as 'f(x)', as 'x' reaches towards positive or negative infinity.
In our example, the polynomial function is of the form
\[f(x) = x^{2}(x - 1).\]
This is a degree 3 polynomial with a leading coefficient of 1. Here's where a little rule of thumb comes in handy: if the degree is odd and the leading coefficient is positive, the ends of the graph go in opposite directions. As 'x' increases without bound (heading toward +infinity), the function will also increase without bound. Conversely, as 'x' decreases without bound (heading toward -infinity), the function in this case will decrease without bound. This creates what we refer to as a 'tail' that extends in the negative direction to the left, and a 'tail' that extends in the positive direction to the right.
The precise behavior can be summarized as:
In our example, the polynomial function is of the form
\[f(x) = x^{2}(x - 1).\]
This is a degree 3 polynomial with a leading coefficient of 1. Here's where a little rule of thumb comes in handy: if the degree is odd and the leading coefficient is positive, the ends of the graph go in opposite directions. As 'x' increases without bound (heading toward +infinity), the function will also increase without bound. Conversely, as 'x' decreases without bound (heading toward -infinity), the function in this case will decrease without bound. This creates what we refer to as a 'tail' that extends in the negative direction to the left, and a 'tail' that extends in the positive direction to the right.
The precise behavior can be summarized as:
- As \(x \rightarrow +\infty\), \(f(x) \rightarrow +\infty\)
- As \(x \rightarrow -\infty\), \(f(x) \rightarrow -\infty\)
X-Intercepts and Zeros of Polynomials
The points where a graph crosses or touches the x-axis are called x-intercepts, which correspond exactly to the zeros or roots of the function. These are the values of 'x' where the function 'f(x)' equals zero. Sifting through the functions for these special numbers tells us exactly where those dips or hops across the x-axis happen.
For the polynomial
\[f(x) = x^{2}(x - 1),\]
we can set \(f(x) = 0\) to find the x-intercepts. Here we get \(x = 0\) with a multiplicity of 2, since the factor \(x^{2}\) suggests the graph will merely touch or graze the axis at this intercept before turning back. Then, there's also \(x = 1\) with a multiplicity of 1, indicating the graph will cross through the axis at this point. Multiplicities give us a hint about the graph's behavior - even multiplicities imply touching the axis, while odd multiplicities imply crossing.
The points on the graph to mark would be
For the polynomial
\[f(x) = x^{2}(x - 1),\]
we can set \(f(x) = 0\) to find the x-intercepts. Here we get \(x = 0\) with a multiplicity of 2, since the factor \(x^{2}\) suggests the graph will merely touch or graze the axis at this intercept before turning back. Then, there's also \(x = 1\) with a multiplicity of 1, indicating the graph will cross through the axis at this point. Multiplicities give us a hint about the graph's behavior - even multiplicities imply touching the axis, while odd multiplicities imply crossing.
The points on the graph to mark would be
- (0, 0), with multiplicity of 2
- (1, 0), with multiplicity of 1
Y-Intercept of Graph
Imagine you're tracing the journey of the graph as it sets off from the origin along the y-axis. The first encounter - that's your y-intercept. It's like the graph's first milestone on its travel up or down the y-axis and occurs where 'x' is zero.
The simple step to find this for our polynomial
\[f(x) = x^{2}(x - 1)\]
is to plug \(x = 0\) into the function. When we substitute, we get \(f(0) = 0^2(0 - 1) = 0\), which pleasantly simplifies to zero. Therefore, the y-intercept of this graph is at the origin (0,0). This single point allows us to anchor our graph on the y-axis and start plotting the rest of the curve with more ease.
The simple step to find this for our polynomial
\[f(x) = x^{2}(x - 1)\]
is to plug \(x = 0\) into the function. When we substitute, we get \(f(0) = 0^2(0 - 1) = 0\), which pleasantly simplifies to zero. Therefore, the y-intercept of this graph is at the origin (0,0). This single point allows us to anchor our graph on the y-axis and start plotting the rest of the curve with more ease.
Intervals of Positivity and Negativity
Ever considered the mood swings of a polynomial function? Well, maybe not in human terms, but recognizing when a function is positive or negative definitely adds character to its graph. These intervals of positivity and negativity are all about identifying where on the x-axis the function smiles (positive) above the axis or frowns (negative) below the axis.
Looking at our polynomial
\[f(x) = x^{2}(x - 1),\]
we utilize the zeros we previously found to break the x-axis into intervals. Here’s how we lay it out:
Looking at our polynomial
\[f(x) = x^{2}(x - 1),\]
we utilize the zeros we previously found to break the x-axis into intervals. Here’s how we lay it out:
- For \(x < 0\), the function is negative, dipping below the x-axis,
- Between the zeros \(x = 0\) and \(x = 1\), the function bounces above the x-axis, having a positive interval because of the even multiplicity at \(x = 0\),
- For \(x > 1\), it ducks back down, resuming negativity beyond the zero at \(x = 1\).
Other exercises in this chapter
Problem 27
Use synthetic division to find the function values. \(f(x)=x^{4}-2 x^{2}+1 ;\) find \(f\left(\frac{1}{2}\right)\)
View solution Problem 27
Determine the end behavior of the function. $$H(x)=-5 x^{4}+3 x^{2}+x-1$$
View solution Problem 28
Solve the rational inequality. $$\frac{x-4}{2 x+1}>0$$
View solution Problem 28
Find all the real zeros of the polynomial. $$P(t)=6 t^{3}-4 t^{2}+3 t-2$$
View solution