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Chapter 2: Problem 65

For each polynomial function given: (a) list each real zero and its multiplicity; (b) determine whether the graph touches or crosses at each \(x\) -intercept; (c) find the \(y\) -intercept and a few points on the graph; (d) determine the end behavior; and (e) sketch the graph. $$f(x)=-(x+2)^{2}(x-1)^{2}$$

Short Answer

Expert verified
Zeros at \( x = -2 \) and \( x = 1 \) (both touch), \( y \)-intercept at -4, graph falls on both ends.

Step by step solution

01

Identify the Real Zeros

The function is given by \( f(x)=-(x+2)^{2}(x-1)^{2} \). The zeros are found by setting the factors equal to zero: \( x+2=0 \) gives \( x=-2 \), and \( x-1=0 \) gives \( x=1 \).
02

Determine Multiplicities and Crossing Behavior

For each zero, check the exponent in the factor. Both zeros have a multiplicity of 2 because the exponents are 2: \( (x+2)^2 \) for \( x=-2 \) and \( (x-1)^2 \) for \( x=1 \). A multiplicity of 2 indicates that the graph touches and turns around at each zero.
03

Find the Y-Intercept

To find the \( y \)-intercept, substitute \( x = 0 \) into the function: \[ f(0) = -(0+2)^2(0-1)^2 = -4. \] This gives the point \( (0, -4) \).
04

Plot Additional Points

Choose a few values of \( x \) to find additional points: \( f(-1) = -1 \), and \( f(2) = -9 \). The points \( (-1, -1) \) and \( (2, -9) \) are found on the graph.
05

Determine End Behavior

The degree of the polynomial is 4, which is even. The leading term of the expanded polynomial is negative, so as \( x \to \pm \infty \), \( f(x) \to -\infty \). This tells us the graph falls on both ends.
06

Sketch the Graph

Using the real zeros, their multiplicities, \( y \)-intercept, additional points, and end behavior, we can sketch the graph. The graph touches the \( x \)-axis at \( x = -2 \) and \( x = 1 \), turns, and falls as \( x \to \pm \infty \).

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Key Concepts

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

Real Zeros
Real zeros are the values of \( x \) that make a polynomial equal to zero. They are crucial because they represent the points where a graph intersects the \( x \)-axis. For the polynomial function \( f(x) = -(x+2)^2(x-1)^2 \), we determine the real zeros by solving \((x+2) = 0\) and \((x-1) = 0\). This yields the zeros \(x = -2\) and \(x = 1\).
  • Finding real zeros helps in sketching the graph since they indicate \(x\)-intercepts.
  • These zeros provide insight into the function's behavior near the \(x\)-axis.
Understanding real zeros allows you to know where the graph changes direction or where it crosses/touches the \(x\)-axis.
Multiplicity
Multiplicity refers to how many times a particular zero appears in a polynomial. It is represented by the exponent of the factor corresponding to the zero. In the function \( f(x) = -(x+2)^2(x-1)^2 \), both zeros \(x = -2\) and \(x = 1\) have a multiplicity of 2.
  • A multiplicity of 2 means the graph touches the \(x\)-axis and bounces back at these points.
  • Odd multiplicities cause the graph to cross the \(x\)-axis.
  • Even multiplicities result in the graph touching and turning without crossing.
Knowing the multiplicity is vital for predicting whether a graph will cross or just touch the \(x\)-axis at its zeros.
X-Intercepts
The \(x\)-intercepts of a graph are the same as the real zeros since they are the points where the graph meets the \(x\)-axis. For \( f(x) = -(x+2)^2(x-1)^2 \), the \(x\)-intercepts are located at \(x = -2\) and \(x = 1\).
  • Each \(x\)-intercept corresponds to a real zero of the polynomial.
  • The behavior at each \(x\)-intercept depends on the zero's multiplicity.
  • For this function, since each zero has a multiplicity of 2, the graph touches and does not cross the \(x\)-axis at these intercepts.
Recognizing \(x\)-intercepts provides a guide to where significant changes in the direction of the graph may occur.
Y-Intercepts
The \(y\)-intercept is the point where the graph crosses the \(y\)-axis. You can find it by evaluating the polynomial function at \(x = 0\). For \( f(x) = -(x+2)^2(x-1)^2 \), this gives: \[ f(0) = -(0+2)^2(0-1)^2 = -4. \]Thus, the \(y\)-intercept is at the point \((0, -4)\).
  • Finding the \(y\)-intercept helps in sketching the exact position of the graph relative to the \(y\)-axis.
  • This is a unique point on the graph where \(x = 0\).
The \(y\)-intercept offers a reference point when finding symmetry and predicting vertical positions along the graph.
End Behavior
End behavior describes how a polynomial function acts as \( x \) approaches positive or negative infinity. It highlights whether the graph rises or falls on either end. In our function \( f(x) = -(x+2)^2(x-1)^2 \), the degree is 4, and the leading coefficient is negative, due to the negative sign before the factors.
  • Since the degree is even and the leading coefficient is negative, the graph falls as \( x \to \pm \infty \).
  • This means both arms of the graph go downwards.
Understanding end behavior helps in depicting the overall direction of the graph, ensuring the sketch reflects the true progression of the polynomial as \( x \) becomes very large or very small.

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