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\(15-26\) Sketch the graph of the polynomial function. Make sure your graph shows all intercepts and exhibits the proper end behavior. $$ P(x)=\frac{1}{12}(x+2)^{2}(x-3)^{2} $$

Short Answer

Expert verified
Graph crosses y-axis at (0, 3), touches x-axis at x = -2 and x = 3, and both ends rise to infinity.

Step by step solution

01

Identify Intercepts

To find the x-intercepts, set the polynomial equal to zero: \( P(x) = \frac{1}{12}(x+2)^{2}(x-3)^{2} = 0 \).This equation is zero when either \((x+2)^{2} = 0\) or \((x-3)^{2} = 0\). Solving these gives us the x-intercepts: \(x = -2\) and \(x = 3\).Since these are even powers, the graph touches the x-axis at these points but does not cross it.The y-intercept is found by evaluating \(P(0)\):\[ P(0) = \frac{1}{12}(0+2)^{2}(0-3)^{2} = \frac{1}{12} imes 4 imes 9 = 3 \].Thus, the y-intercept is at \( (0, 3) \).
02

Determine End Behavior

The leading term's power of the polynomial is \((x^4)\), indicating this is a quartic (fourth-degree) polynomial.Quartic polynomials have the same end behavior as quadratics. Because the leading coefficient \( \frac{1}{12} \) is positive, both ends of the graph rise to positive infinity as \(x\to \pm\infty\).
03

Sketch the Graph

Using the intercepts and end behavior, sketch the graph.The graph touches the x-axis at \(x = -2\) and \(x = 3\), and rises to positive infinity as \(x\) approaches both positive and negative infinity.The y-intercept is at \((0,3)\) so the graph should pass through this point.Draw the graph starting at the bottom left, touching but not crossing at \(x=-2\), through \((0,3)\), touching but not crossing at \(x=3\), and rising towards positive infinity.

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Key Concepts

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

X-intercepts
X-intercepts are the points where the graph of a polynomial crosses or touches the x-axis. These occur where the polynomial equals zero. In our quartic polynomial \( P(x) = \frac{1}{12}(x+2)^{2}(x-3)^{2} \), the x-intercepts are identified by solving \( P(x) = 0 \). This requires that either \((x+2)^2 = 0\) or \((x-3)^2 = 0\). Solving these equations yields the x-intercepts at \(x = -2\) and \(x = 3\).

However, because the factors \((x+2)^2\) and \((x-3)^2\) are squared (even powers), these x-intercepts indicate that the graph touches the x-axis at these points but does not actually cross it. In the case of even powers, the behavior at these intercepts is called a "bounce," where the graph approaches the axis, makes contact, and then reverses direction.
Y-intercept
The y-intercept of a polynomial is the point where the graph crosses the y-axis. This occurs when \(x = 0\). To find the y-intercept of \( P(x) = \frac{1}{12}(x+2)^2(x-3)^2 \), substitute \(x = 0\) into the polynomial. This calculation looks like:
  • \[ P(0) = \frac{1}{12}(0+2)^2(0-3)^2 \]
  • \[ P(0) = \frac{1}{12} \times 4 \times 9 \]
  • \[ P(0) = 3 \]
Therefore, the y-intercept is at the point \((0, 3)\). This means that the graph will cross the y-axis at this height, allowing us to plot this vital point on our graph. The y-intercept provides a fixed point the graph must pass through, aiding in visualizing the polynomial's shape.
End Behavior
End behavior describes how a polynomial graph behaves as \(x\) approaches positive or negative infinity. For the polynomial \( P(x) = \frac{1}{12}(x+2)^2(x-3)^2 \), the highest power of \(x\) is \(x^4\), confirming it as a quartic polynomial.

The end behavior of quartic polynomials mimics the behavior of quadratic ones due to their even degree. In this case, since the leading coefficient \(\frac{1}{12} > 0\), both ends of the polynomial graph rise to positive infinity. This behavior is formally described as \( P(x) \to +\infty \) as \( x \to \pm\infty \). The graph begins in the lower left quadrant, rises through important points, and heads towards the upper right and left quadrants, regardless of how the middle behaves.
Quartic Polynomial
A quartic polynomial is a polynomial of degree four. The standard form of a quartic polynomial is \( ax^4 + bx^3 + cx^2 + dx + e \), where \(a eq 0\). In our example, \( P(x) = \frac{1}{12}(x+2)^2(x-3)^2 \), the polynomial equals \(\frac{1}{12}(x^2 + 4x + 4)(x^2 - 6x + 9)\). If expanded, it would reveal the true quartic form.

Quartic polynomials can have up to four real roots (including multiplicities) and can exhibit complex but symmetric shapes due to the even degree. They can touch or cross the x-axis up to four times, depending on the nature of their roots. This specific quartic function has the peculiar property of touching but not crossing the x-axis at its double roots, leading to a distinctive "bounce" at these intercepts, adding to the rich and intuitive visualization when graphing polynomial functions.

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Most popular questions from this chapter

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