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Chapter 3: Problem 11

Sketch the graph of the polynomial function. Make sure your graph shows all intercepts and exhibits the proper end behavior. $$P(x)=(x-1)(x+2)$$

Short Answer

Expert verified
Intercepts: (1, 0), (-2, 0), and (0, -2). Graph is a parabola opening upwards.

Step by step solution

01

Identify the Intercepts

The polynomial function is given by \(P(x)=(x-1)(x+2)\). To find the x-intercepts, set \(P(x) = 0\). Solving \((x-1)(x+2) = 0\), we have two solutions: \(x = 1\) and \(x = -2\). These are the x-intercepts of the function. The y-intercept occurs when \(x = 0\), so calculate \(P(0) = (0-1)(0+2) = -2\). Thus, the y-intercept is \((0, -2)\).
02

Determine End Behavior

Consider the degree and leading coefficient of the polynomial. The expression is already factored as \((x-1)(x+2)\), which multiplies to \(x^2 + x - 2\). Therefore, it is a quadratic polynomial with a positive leading coefficient. As \(x\rightarrow\infty\), \(P(x)\rightarrow\infty\) and as \(x\rightarrow -\infty\), \(P(x)\rightarrow\infty\). The graph opens upwards because the leading coefficient is positive.
03

Sketch the Graph

Plot the intercepts on a coordinate plane: intercepts are \((1, 0)\), \((-2, 0)\), and \((0, -2)\). Knowing the end behavior and that the graph is a parabola, sketch a curve opening upwards, passing through these intercepts. Ensure the tails of the parabola go upwards, confirming the end behavior deduced earlier.

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Key Concepts

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

x-intercepts
The x-intercepts of a polynomial function are the points where the graph of the function crosses the x-axis. These are crucial for graphing, as they depict where the function equals zero.
To find the x-intercepts of the polynomial function given by \( P(x)=(x-1)(x+2) \), we solve the equation \( P(x) = 0 \).
This involves finding the values of \( x \) that make \( P(x) \) equal to zero.
  • First, set each factor of the polynomial to zero: \((x-1)=0\) and \((x+2)=0\).
  • Solving \(x-1=0\) gives \(x=1\).
  • Solving \(x+2=0\) gives \(x=-2\).
Thus, the x-intercepts are \( (1, 0) \) and \( (-2, 0) \). These indicate that at \( x=1 \) and \( x=-2 \), the graph crosses the x-axis.
y-intercepts
The y-intercept of a polynomial function is where the graph crosses the y-axis. This occurs when \( x = 0 \).
It represents the output of the polynomial when there is no input variable, which is a vital point for plotting the graph.
To find the y-intercept of \( P(x)=(x-1)(x+2) \), we calculate \( P(0) \) by substituting 0 in place of \( x \):
  • \( P(0) = (0 - 1)(0 + 2) = -1 \times 2 = -2 \).
So, the y-intercept is \((0, -2)\).
This means the graph will intersect the y-axis at the point where y is -2, serving as another anchor point for sketching the curve.
end behavior
End behavior in polynomial functions describes how the function behaves as \( x \) moves towards positive infinity \(( +\infty \)) or negative infinity \(( -\infty \)).
This is determined by the leading term of the polynomial when it is fully expanded. For the polynomial \( P(x)=(x-1)(x+2) \), it expands to \( x^2 + x - 2 \), with a leading term of \( x^2 \).
Let's understand the implications:
  • The degree of the polynomial (2, in this case) indicates it is quadratic.
  • The leading coefficient of \(x^2\) is positive.
Given this, as \( x \to \infty \), \( P(x) \to \infty \), and as \( x \to -\infty \), \( P(x) \to \infty \).
This means the ends of the graph will rise upward, creating the characteristic 'U' shape of a parabola opening upwards.
quadratic polynomial
A quadratic polynomial is a polynomial of degree 2, characterized by the general form \( ax^2 + bx + c \).
These polynomials graph as parabolas, which may open upwards or downwards depending on the leading coefficient. For \( P(x)=(x-1)(x+2) \), upon expansion, we have \( P(x)=x^2+x-2 \).
  • The polynomial has a leading term of \( x^2 \), confirming a degree of 2.
  • The leading coefficient \( a \) is 1, which is positive, indicating the parabola opens upwards.
Main characteristics include:
  • Vertex: The highest or lowest point of the parabola. Not calculated directly here, but lies halfway between the roots \( x = 1 \) and \( x = -2 \).
  • Symmetry: Quadratic graphs are symmetric about a vertical line passing through the vertex.
Understanding these properties helps predict the overall shape and behavior of the quadratic polynomial's graph.

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Most popular questions from this chapter

Use transformations of the graph of \(y=\frac{1}{x}\) to graph the rational function, as in Example 2. $$s(x)=\frac{-2}{x-2}$$

Find the slant asymptote, the vertical asymptotes, and sketch a graph of the function. $$r(x)=\frac{3 x-x^{2}}{2 x-2}$$

A polynomial \(P\) is given. (a) Find all the real zeros of \(P\). (b) Sketch the graph of \(P\). $$P(x)=2 x^{3}-7 x^{2}+4 x+4$$

Show that the given values for \(a\) and \(b\) are lower and upper bounds for the real zeros of the polynomial. $$P(x)=8 x^{3}+10 x^{2}-39 x+9 ; \quad a=-3, b=2$$

Graph the rational function and find all vertical asymptotes, \(x\)- and \(y\)-intercepts, and local extrema, correct to the nearest decimal. Then use long division to find a polynomial that has the same end behavior as the rational function, and graph both functions in a sufficiently large viewing rectangle to verify that the end behaviors of the polynomial and the rational function are the same. $$y=\frac{x^{4}-3 x^{3}+x^{2}-3 x+3}{x^{2}-3 x}$$
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