/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 87 Sketching the Graph of a Polynom... [FREE SOLUTION] | 91Ó°ÊÓ

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Sketching the Graph of a Polynomial Function, sketch the graph of the function by (a) applying the Leading Coefficient Test, (b) finding the real zeros of the polynomial, (c) plotting sufficient solution points, and (d) drawing a continuous curve through the points. $$ g(t)=-\frac{1}{4}(t-2)^{2}(t+2)^{2} $$

Short Answer

Expert verified
The graph of this polynomial function begins and ends falling to the left and right. It crosses the x-axis at points -2 and 2 and has the shape of a downward opening 'W'.

Step by step solution

01

Apply the Leading Coefficient Test

The Leading Coefficient Test allows us to determine the direction of the polynomial. The leading coefficient for \(g(t)=-\frac{1}{4}(t-2)^{2}(t+2)^{2}\) is -1/4, which is negative. The degree of the polynomial is 4, which is even. Therefore, if the leading coefficient is negative and the degree is even, the graph will fall to the left and right.
02

Finding The Real Zeros

The real zeros of \(g(t)\) come from the roots of \((t-2)^{2}(t+2)^{2}=0\). Those zeros are \(t=-2, 2\) (from the equation \((t-2)(t+2)=0\)).
03

Plotting sufficient solution points

Now we choose points between and outside the zeros to determine the behavior of the polynomial. For instance, let us evaluate \(g(t)\) for \(t=-3,-1,0,1,3\). This will give us sufficient solution points to plot our graph.
04

Drawing a continuous curve

Plot these points and the identified zeros on the number line. We get a point at \(-2\), \(-1\), \(0\), \(1\), \(2\), and \(3\). Taking these points together, and remembering the Leading Coefficient Test results, draw a continuous curve through these points that rises and falls appropriately. This gives us the graph of our polynomial function.

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Key Concepts

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

Leading Coefficient Test
When sketching the graph of a polynomial function, the Leading Coefficient Test is a valuable tool that guides us about the end behavior of the graph. The leading coefficient is the coefficient of the term with the highest power, which tells us whether the graph will rise or fall as it moves towards infinity or negative infinity. In our example, the function is given by

\[ g(t)=-\frac{1}{4}(t-2)^{2}(t+2)^{2}\]

Since the leading coefficient is negative and the highest power of the polynomial (the degree) is even, we can infer that the ends of this graph will fall in both directions. This means as we move towards extremities on both sides of the x-axis, the graph will head downwards. This sets the groundwork for how we start to imagine the shape of our function.
Real Zeros of Polynomial
The real zeros of a polynomial are the values of the variable at which the polynomial equals zero. For the polynomial given in our exercise, these zeros can be found by setting the function equal to zero and solving for 't'.

\[ g(t)=-\frac{1}{4}(t-2)^{2}(t+2)^{2} = 0\]

Here, we have \((t-2)^2(t+2)^2=0\), which means our real zeros are at \( t = 2 \) and \( t = -2 \). These zeros are crucial for plotting the graph because they represent the x-intercepts, where our graph will cross or touch the x-axis. Knowing where these points lie helps shape the graph correctly.
Plotting Solution Points
To depict the behavior of the polynomial between and beyond the zeros, we select strategic points and calculate their corresponding output values. This step is like setting up a frame to support the overall structure of the graph. By choosing \( t = -3, -1, 0, 1, \) and \( 3 \), we use these to determine the graph's curvature and direction. After calculating the value of the polynomial at these points,

\[ g(-3), g(-1), g(0), g(1), g(3) \]

we plot them on a Cartesian plane. We end up with a set of dots which, when connected, will begin to suggest the visual form of our polynomial's graph. It's like connecting the dots but with a mathematical purpose.
Continuous Curve Graphing
With the zeros and additional solution points plotted, the final touch is to draw a continuous, smooth curve through these points. The masterpiece of our function's graph should be a snapshot of the function at all points. Since polynomial functions are continuous and smooth, there are no breaks, jumps, or sharp turns in the graph. The direction of our curve is informed by our analysis - the Leading Coefficient Test sets the end behavior, and the plotted points guide the detail. By gracefully arching through our zeros and bending to meet our solution points, we create a visual model of our function that is as accurate as it is helpful for understanding the behavior of the polynomial across the entire number line.

In our specific instance, the curve will dip downward at the ends following our Leading Coefficient Test outcome, and arch upward between \( t = -2 \) and \( t = 2 \), matching the parabolic shape suggested by the quadruplicity of the real zeros.

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Most popular questions from this chapter

Think About It For each function, identify the degree of the function and whether the degree of the function is even or odd. Identify the leading coefficient and whether the leading coefficient is positive or negative. Use a grading coefficient graph each functive. Describe the relationship between the degree of the function and the sign of the leading coefficient of the function and the right-hand and left-hand behavior of the graph of the function. (a) \(f(x)=x^{3}-2 x^{2}-x+1\) (b) \(f(x)=2 x^{5}+2 x^{2}-5 x+1\) (c) \(f(x)=-2 x^{5}-x^{2}+5 x+3\) (d) \(f(x)=-x^{3}+5 x-2\) (e) \(f(x)=2 x^{2}+3 x-4\) (f) \(f(x)=x^{4}-3 x^{2}+2 x-1\) (g) \(f(x)=x^{2}+3 x+2\)

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Comparing Models If \(\$ 1\) is invested over a 10 -year period, then the balance \(A,\) where \(t\) represents the time in years, is given by \(A=1+0.075[t]\) or \(A=e^{0.07 t}\) depending on whether the interest is simple interest at 7\(\frac{1}{2} \%\) or continuous compound interest at 7\(\% .\) Graph each function on the same set of axes. Which grows at a greater rate? (Remember that \([t]\) is the greatest integer function discussed in Section \(1.6 . )\)

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