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Chapter 4: Problem 60

(a) Factor the expression \(4 x^{2}-x^{4}\). Then use the techniques explained in this section to graph the function defined by \(y=4 x^{2}-x^{4}\). (b) Find the coordinates of the turning points. Hint: As in previous sections, use the substitution \(x^{2}=t\).

Short Answer

Expert verified
The function has turning points at \((\sqrt{2}, 4)\) and \((-\sqrt{2}, 4)\).

Step by step solution

01

Factor the Expression

The expression to factor is \( 4x^2 - x^4 \). Begin by recognizing a common structure: rewrite it as \( -x^4 + 4x^2 \). Notice this can be further factored by taking out \(-x^2\) common factor: \( -x^2(x^2 - 4) \). Finally, \( x^2 - 4 \) is a difference of squares, which can be factored as \((x - 2)(x + 2)\). Therefore, the complete factored form is \(-x^2(x - 2)(x + 2)\).
02

Graph the Function

Using the factored form \(-x^2(x-2)(x+2)\), note the roots of the function are \(x = 0, x = 2, x = -2\). Since the leading coefficient of the original \(x^4\) term is negative, the end behaviors of the function imply that as \(x \to \pm \infty\), \(y \to -\infty\). These features, along with the symmetry about the y-axis (since all exponents in the original equation are even), help sketch the graph.
03

Substitute to Find Turning Points

Substitute \(t = x^2\) into \(y = 4x^2 - x^4\), transforming the equation to \(y = 4t - t^2\). This is a quadratic function \(-t^2 + 4t\), which can be rewritten as \(-1(t^2 - 4t)\). Completing the square gives \((t - 2)^2 - 4\). The vertex form \(-1((t - 2)^2 - 4)\) indicates a maximum at \(t=2\). Re-substitute \(t = x^2\), solving \(x^2=2\), yields \(x = \pm \sqrt{2}\).
04

Calculate Function Value at Turning Points

Using the x-values from Step 3, calculate \(y\) for \(x = \sqrt{2}\) and \(x = -\sqrt{2}\): \(y = 4(\sqrt{2})^2 - (\sqrt{2})^4\). Simplifying, \(y = 8 - 4 = 4\). Thus, the coordinates of the turning points are \((\sqrt{2}, 4)\) and \((-\sqrt{2}, 4)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Turning Points
Turning points are special points on the graph of a function where the direction changes from increasing to decreasing or vice versa. These points are crucial as they help in understanding the behavior of a polynomial function and finding these can make graphing simpler.

To find turning points, specifically for the function defined by \[ y = 4x^2 - x^4 \], we can utilize a substitution technique. By substituting \( x^2 = t \), we transform the quartic equation into a quadratic one: \( y = 4t - t^2 \). This quadratic format allows us to utilize the vertex form to easily identify turning points.

In the vertex form \( y = -(t - 2)^2 + 4 \), the turning point (or vertex) is observed at \( t = 2 \), which translates back to \( x = \pm \sqrt{2} \) when reversing the substitution. Thus, these x-values give us turning points at coordinates \( (\sqrt{2}, 4) \) and \( (-\sqrt{2}, 4) \).

These turning points indicate maximums because the coefficient of \( t^2 \) is negative, confirming the parabola opens downwards, providing clear guidance on graphing the function.
Difference of Squares
The difference of squares is a very useful factoring technique that comes in handy when simplifying quadratic expressions. It applies to expressions in the form of \( a^2 - b^2 \), where it can be factored into \((a - b)(a + b)\).

In the expression to factor, \( 4x^2 - x^4 \), the rewritten form is \( -x^2(x^2 - 4) \), where \( x^2 - 4 \) can be easily recognized as a difference of squares because \( 4 \) is \( 2^2 \).

This difference of squares factors into \( (x - 2)(x + 2) \), thus simplifying the process of solving and graphing the polynomial. Factoring using the difference of squares helps in identifying the roots or zeros of the function, laying a strong foundation for graphing and further analysis.
Graphing Functions
Graphing functions involves understanding their general shape and key features like roots, turning points, and end behavior. With the function \( y = 4x^2 - x^4 \), knowing it can be expressed as \( -x^2(x - 2)(x + 2) \) reveals useful information for graphing.

The roots of this polynomial are directly obtained from the factored form: \( x = 0, x = 2, \text{and} x = -2 \), indicating where the graph will intersect the x-axis.

The turning points found earlier at \( (\sqrt{2}, 4) \) and \( (-\sqrt{2}, 4) \) show where the graph reaches its maximum height and starts to descend.

Moreover, being a degree 4 polynomial with a leading negative coefficient indicates the end behavior as \( x \to \pm \infty, y \to -\infty \). The graph is symmetric about the y-axis, evident from the original expression's even powers, guiding how we visualize the function鈥檚 graph.
  • Primary roots: \( x = 0, x = 2, x = -2 \)
  • Turning points: \( (\sqrt{2}, 4) \), and \( (-\sqrt{2}, 4) \)
  • End behavior trends towards \(-\infty \) as \( x \to \pm \infty \)
These features help in sketching an accurate graph of the function.

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Most popular questions from this chapter

Determine the inputs that yield the minimum values for each function. Compute the minimum value in each case. (a) \(f(x)=\sqrt{x^{2}-6 x+73}\) (b) \(g(x)=\sqrt[3]{x^{2}-6 x+73}\) (c) \(h(x)=x^{4}-6 x^{2}+73\)

(a) Use a graphing utility to graph the parabolas \(y=x^{2}-2 x, y=x^{2}+2 x, y=x^{2}-3 x,\) and \(y=x^{2}+3 x\) Check visually that, in each case, the vertex of the parabola appears to lie on the curve \(y=-x^{2}\) (b) Prove that for all real numbers \(k\), the vertex of the parabola \(y=x^{2}+k x\) lies on the curve \(y=-x^{2}\)

Let \(A B C\) be a triangle with vertices \(A(-3,0), B(0,6)\) \(C(3,0) .\) Let \(P\) be a point on the line segment that joins \(B\) to the origin. Find the position of \(P\) that minimizes the sum of the distances between \(P\) and the vertices.

Let \(f(x)=\left(x^{5}+1\right) / x^{2}\) (a) Graph the function \(f\) using a viewing rectangle that extends from -4 to 4 in the \(x\) -direction and from -8 to 8 in the \(y\) -direction. (b) Add the graph of the curve \(y=x^{3}\) to your picture in part (a). Note that as \(|x|\) increases (that is, as \(x\) moves away from the origin), the graph of \(f\) looks more and more like the curve \(y=x^{3} .\) For additional perspective, first change the viewing rectangle so that \(y\) extends from -20 to \(20 .\) (Retain the \(x\) -settings for the moment.) Describe what you see. Next, adjust the viewing rectangle so that \(x\) extends from -10 to 10 and \(y\) extends from -100 to \(100 .\) Summarize your observations. (c) In the text we said that a line is an asymptote for a curve if the distance between the line and the curve approaches zero as we move further and further out along the curve. The work in part (b) illustrates that a curve can behave like an asymptote for another curve. In particular, part (b) illustrates that the distance between the curve \(y=x^{3}\) and the graph of the given function \(f\) approaches zero as we move further and further out along the graph of \(f .\) That is, the curve \(y=x^{3}\) is an "asymptote" for the graph of the given function \(f\). Complete the following two tables for a numerical perspective on this. In the tables, \(d\) denotes the vertical distance between the curve \(y=x^{3}\) and the graph of \(f:\) $$ d=\left|\frac{x^{5}+1}{x^{2}}-x^{3}\right| $$ $$\begin{array}{llllll} \hline x & 5 & 10 & 50 & 100 & 500 \\ \hline d & & & & \\ \hline & & & & \\ \hline x & -5 & -10 & -50 & -100 & -500 \\ \hline d & & & & \\ \hline \end{array}$$ (d) Parts (b) and (c) have provided both a graphical and a numerical perspective. For an algebraic perspective that ties together the previous results, verify the following identity, and then use it to explain why the results in parts (b) and (c) were inevitable: $$ \frac{x^{5}+1}{x^{2}}=x^{3}+\frac{1}{x^{2}} $$

Sketch the graph of each rational function. Specify the intercepts and the asymptotes. $$y=1 /(x+2)^{3}$$
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