Problem 55
Question
A. Use the Leading Coefficient Test to determine the graph's end behavior. B. Find the \(x\) -intercepts. State whether the graph crosses the \(x\) -axis, or touches the \(x\) -axis and turns around, at each intercept. C. Find the \(y\) -intercept. D. Determine whether the graph has \(y\) -axis symmetry, origin symmetry, or neither. E. If necessary, find a few additional points and graph the function. Use the maximum number of uning points to check whether it is drawn correctly. $$ f(x)=-3(x-1)^{2}\left(x^{2}-4\right) $$
Step-by-Step Solution
Verified Answer
The end behavior of the function is that as x approaches negative or positive infinity, f(x) approaches positive infinity. The x-intercepts are x = 1, 2, -2 and the graph crosses the x-axis at these points. The y-intercept is (0,-12) and the graph has no symmetry about y-axis or the origin.
1Step 1: Determine the end behavior
The degree of the given polynomial is 4, which is even. The leading coefficient, -3, is negative. Therefore the end behaviors of the graph can be determined as follows: As \(x \rightarrow -\infty\), \(f(x) \rightarrow +\infty\) and as \(x \rightarrow +\infty\), \(f(x) \rightarrow +\infty\).
2Step 2: Find the x-intercepts
Setting the polynomial to zero and solving for x will give the x-intercepts. Therefore -3(x-1)^{2}(x^{2}-4) = 0 implies x = 1, 2, -2, as x-intercepts. The graph will cross x-axis at these points, because the powers of the corresponding factors are even.
3Step 3: Find the y-intercept
Setting x = 0 in the function will give the y-intercept. Therefore, f(0) = -3(0-1)^{2}(0^{2}-4) = -12. Therefore, the y-intercept is (0, -12).
4Step 4: Determine the symmetry
The function is not symmetric about y-axis or origin. This can be inferred from the function as replacing x by -x does not result in the original function (not symmetric about y-axis) or the negative of the function (not symmetric about origin).
5Step 5: Plot additional points and check
Following the above steps, the main characteristics of the graph are obtained. Now, a few more points may be plotted for completeness of the graph. It can also be verified that the graph touches or crosses the x-axis at the x-intercepts and has the correct end behavior.
Key Concepts
Leading coefficient testx-interceptsy-interceptsSymmetry in graphs
Leading coefficient test
The leading coefficient test is a helpful tool to predict the end behavior of polynomial functions' graphs. To perform the leading coefficient test, you'll first need to identify two main features of the polynomial: its degree and its leading coefficient.
The degree of the polynomial is the highest power of the variable in the expression. In our function \(f(x) = -3(x-1)^{2}(x^{2}-4)\), the polynomial is expanded into a degree of 4, because when multiplied out, the highest power of \(x\) will be 4. An even degree tells us that the arms of the graph either both point upward or both point downward.
The coefficient of the term with the highest degree is the leading coefficient, here \-3\. A negative leading coefficient of an even-degree polynomial indicates that both ends of the graph will move towards negative infinity as you move away from the origin. This implies that the graph starts high as \(x\)\rightarrow \(-\infty\) and descends low as \(x\rightarrow \infty\).
The degree of the polynomial is the highest power of the variable in the expression. In our function \(f(x) = -3(x-1)^{2}(x^{2}-4)\), the polynomial is expanded into a degree of 4, because when multiplied out, the highest power of \(x\) will be 4. An even degree tells us that the arms of the graph either both point upward or both point downward.
The coefficient of the term with the highest degree is the leading coefficient, here \-3\. A negative leading coefficient of an even-degree polynomial indicates that both ends of the graph will move towards negative infinity as you move away from the origin. This implies that the graph starts high as \(x\)\rightarrow \(-\infty\) and descends low as \(x\rightarrow \infty\).
- Degree is even, so both ends go in the same direction.
- Leading coefficient is negative, so both ends go downward.
x-intercepts
Finding the \(x\)-intercepts involves setting the polynomial function equal to zero and solving for \(x\). These points tell us where the graph of the function will intersect the \(x\)-axis. In the function \(f(x)=-3(x-1)^{2}(x^{2}-4)\), set \(f(x)=0\).
This simplifies to two factors: \(x-1\) and \(x^{2}-4\), which gives us the zeroes \(x=1\), \(x=2\), and \(x=-2\). These are the \(x\)-intercepts of the graph:
This simplifies to two factors: \(x-1\) and \(x^{2}-4\), which gives us the zeroes \(x=1\), \(x=2\), and \(x=-2\). These are the \(x\)-intercepts of the graph:
- At \(x=1\), \(x=2\), and \(x=-2\), the graph will cross the \(x\)-axis.
y-intercepts
To find the \(y\)-intercept of a polynomial function, you substitute \(x=0\) into the equation. This calculation gives the point where the graph crosses the \(y\)-axis. For our function \(f(x)=-3(x-1)^{2}(x^{2}-4)\), substitute \(x=0\) resulting in:
\[f(0) = -3(0-1)^{2}(0^{2}-4) = -12\]
Thus, the \(y\)-intercept is at the point \((0, -12)\). The graph will intersect the \(y\)-axis at this point, providing a reference for plotting additional points or sketching the graph.
\[f(0) = -3(0-1)^{2}(0^{2}-4) = -12\]
Thus, the \(y\)-intercept is at the point \((0, -12)\). The graph will intersect the \(y\)-axis at this point, providing a reference for plotting additional points or sketching the graph.
Symmetry in graphs
Symmetry in graphs can be a powerful tool in understanding the behavior of a function's graph. In particular, we're often interested in \(y\)-axis symmetry (even functions) and origin symmetry (odd functions).
To check for \(y\)-axis symmetry, replace \(x\) with \(-x\) in the function and see if the expression remains unchanged. For origin symmetry, when \(x\) is replaced, the result should be the negative of the original function.
In the case of \(f(x)=-3(x-1)^{2}(x^{2}-4)\), neither of these conditions holds:
To check for \(y\)-axis symmetry, replace \(x\) with \(-x\) in the function and see if the expression remains unchanged. For origin symmetry, when \(x\) is replaced, the result should be the negative of the original function.
In the case of \(f(x)=-3(x-1)^{2}(x^{2}-4)\), neither of these conditions holds:
- Substituting \(x = -x\) does not result in the same function, implying no \(y\)-axis symmetry.
- Nor does it yield \(-f(x)\), meaning there is no symmetry about the origin.
Other exercises in this chapter
Problem 55
Solve each rational inequality in Exercises \(43-60\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ \frac
View solution Problem 55
The heat generated by a stove element varies directly as the square of the voltage and inversely as the resistance. If the voltage remains constant, what needs
View solution Problem 55
During the 1980 s, the controversial economist Arthur Laffer promoted the idea that tax increases lead to a reduction in government revenue. Called supply- side
View solution Problem 56
In Exercises \(45-56,\) use transformations of \(f(x)-\frac{1}{x}\) or \(f(x)-\frac{1}{x^{2}}\) to graph each rational function. $$h(x)=\frac{1}{(x-3)^{2}}+2$$
View solution