Problem 45 - College Algebra

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. $$ f(x)=x^{4}+16 x^{2} $$

Step-by-Step Solution

Answer

The end behavior of the function is such that as \(x \rightarrow -\infty, f(x) \rightarrow +\infty\) and as \(x \rightarrow +\infty, f(x) \rightarrow +\infty\). The graph touches x-axis at x=0. The y-intercept is at y=0. The graph has y-axis symmetry and the final graph of the function would reflect this and the other findings.

Step 1: Determine the End Behaviour

The degree of the polynomial is 4, which is an even number, and its leading coefficient is 1, which is positive. By the Leading Coefficient Test, because the exponent on the leading term is even, the function will end in the same direction. Since the leading coefficient is positive, the function will rise to the right. Therefore, as \(x \rightarrow -\infty, f(x) \rightarrow +\infty\) and as \(x \rightarrow +\infty, f(x) \rightarrow +\infty\).

Step 2: Find the x-intercepts

To find the x-intercepts, set \( f(x) = 0 \). So, we get that \(x^{4}+16x^{2}=0\). This equation simplifies into \(x^2(x^2+16) = 0\). The solutions are \(x = 0\) and for \(x^2+16=0\), there are no real solutions as \(x^2+16>0\) for all \(x\). So, the graph touches the x-axis at \(x = 0\).

Step 3: Find the y-intercept

To find the y-intercept, set \( x = 0 \) in the given equation. We get that \(f(0) = (0)^{4}+16(0)^{2} = 0\). So, the y-intercept is \(0\).

Step 4: Determine the Symmetry

The given function is \( f(x) = x^{4}+16x^{2} \) which is equivalent to \( f(-x) = (-x)^{4}+16(-x)^{2} = x^{4}+16x^{2} \). Since \( f(-x) = f(x) \), the graph of the function has y-axis symmetry. There is no origin symmetry as \( f(-x) \neq -f(x) \).

Step 5: Graph the Function

Sketch the graph using information obtained from the previous steps. The graph touches the x-axis at x=0 and rises in both directions. The graph also has a y-axis symmetry.

Key Concepts

Leading Coefficient Test

The leading coefficient test is a useful tool to predict how the ends of a polynomial function's graph behave. For any polynomial, the leading coefficient is simply the coefficient in front of the term with the highest degree. - If the degree of the polynomial is even, the ends of the graph will go in the same direction.- If the degree is odd, the ends will go in opposite directions.In our example, the polynomial function is given by \( f(x) = x^4 + 16x^2 \). The degree is 4, an even number, and the leading coefficient is 1, a positive number. Therefore, according to the leading coefficient test:- both arms of the graph will rise, pointing upwards. - As \( x \to -\infty \), \( f(x) \to +\infty \) and similarly, as \( x \to +\infty \), \( f(x) \to +\infty \). This understanding is crucial for sketching and analyzing the end behavior of polynomial functions.

End Behavior

The end behavior of a polynomial function describes what happens to the function's values as the input \( x \) either increases or decreases without bound. In other words, we are interested in knowing the direction in which the graph heads as we move far to the right or far to the left.- In the polynomial \( f(x) = x^4 + 16x^2 \), the leading term is \( x^4 \), which determines the end behavior.- Since the leading coefficient is positive and the degree is even, the graph rises on both sides.To state formally, we say that as \( x \) approaches infinity (\( x \to +\infty \)), \( f(x) \to +\infty \) and as \( x \to -\infty \), \( f(x) \to +\infty \) as well. Understanding end behavior helps us to anticipate the general shape of the graph even without detailed plotting.

x-intercepts

The \(x\)-intercepts of a polynomial function are the points where the graph crosses the \(x\)-axis. They can be found by setting the function equal to zero and solving for \(x\).In the example function \( f(x) = x^4 + 16x^2 \), we find:- Setting \( f(x) = 0 \) gives \( x^4 + 16x^2 = 0 \).- Factoring out \( x^2 \), we have \( x^2 (x^2 + 16) = 0 \).Only \( x = 0 \) leads to a real solution since \( x^2 + 16 = 0 \) does not produce real roots. Thus, the only \(x\)-intercept is at \( x = 0 \). The function touches the \(x\)-axis at this point, but does not cross it, meaning the graph turns back around instead of passing through.

y-intercept

The y-intercepts occur where the graph crosses the y-axis. This happens when all \( x \) values are set to zero in the function. It is essentially looking for \( f(0) \).For \( f(x) = x^4 + 16x^2 \), let's determine the y-intercept:- Set \( x = 0 \) which gives \( f(0) = 0^4 + 16 \, \cdot \, 0^2 = 0 \).Therefore, the \( y \)-intercept of the function is at the origin, (0, 0). This point is usually the starting point in sketches and often is the point where \( x \) and \( y \) interact directly on the graph.

Symmetry of Functions

Symmetry in functions aids in understanding their shape. Symmetrical functions can be easily recognized by certain characteristics when tested for symmetry.1. **Y-axis Symmetry** (Even Function): - A function \( f(x) \) is symmetrical about the \( y \)-axis if replacing \( x \) with \( -x \) yields the same function, \( f(x) = f(-x) \). - Here, \( f(x) = x^4 + 16x^2 \) and \( f(-x) = (-x)^4 + 16(-x)^2 = x^4 + 16x^2 \) confirms y-axis symmetry.2. **Origin Symmetry** (Odd Function): - A function \( f(x) \) is symmetrical about the origin if \( f(-x) = -f(x) \), but this is not the case here as \( f(-x) eq -f(x) \).In conclusion, our function \( f(x) = x^4 + 16x^2 \) demonstrates y-axis symmetry and lacks origin symmetry, making it an even function. Recognizing these properties helps simplify graphing and deepens understanding of function behavior.