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In Exercises 75 - 88, sketch the graph of the function by (a) applying the Leading Coefficient Test, (b) finding the zeros of the polynomial, (c) plotting sufficient solution points, and(d) drawing a continuous curve through the points. \( g(x) = -x^2 + 10x - 16 \)

Short Answer

Expert verified
The graph of the function \( g(x) = -x^2 + 10x - 16 \) will be a downward opening parabola with zeros at x = 2 and x = 8. The function falls to the left and right.

Step by step solution

01

Applying the Leading Coefficient Test

The leading coefficient of the function \( g(x) = -x^2 + 10x - 16 \) is -1 and the degree of the polynomial is 2, which is even. According to the Leading Coefficient Test, if the leading coefficient is negative and the degree of the polynomial is even, the graph falls to the left and falls to the right.
02

Finding the Zeros of the Polynomial

Set \( g(x) = -x^2 + 10x - 16 \) equal to zero and solve for x. \n 0 = -x^2 + 10x - 16 \n which simplifies to \n 0 = -(x - 2)(x - 8) \n Here, the solutions are x = 2, x = 8.
03

Plotting Sufficient Solution Points

We will pick a point to the left of x = 2, between x = 2 and x = 8, and to the right of x = 8 to have a more accurate plot. For example, pick x = 1, x = 5, and x = 9 and find the respective y values for these x values.
04

Drawing a Continuous Curve

Using these points and the zeros x = 2, x = 8, and knowing that the end behavior falls to the left and right due to the degree and leading coefficient, we can draw a continuous curve through the points.

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Key Concepts

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

Leading Coefficient Test
The Leading Coefficient Test is a useful tool in graphing polynomials. It helps us determine how the graph behaves at its ends, or as we say in math, the 'end behavior'. When you examine a polynomial, like the quadratic function \( g(x) = -x^2 + 10x - 16 \), the leading coefficient is the number in front of the highest power of \( x \). In this case, it's -1. The degree of the polynomial is the highest power, which here is 2, making it an even polynomial.
For even-degree polynomials, if the leading coefficient is negative, the graph falls as \( x \) approaches both positive and negative infinity. In simpler words, the arms of the graph point downward on both sides. So, for our function \( g(x) \), you can expect that the graph will fall to the left and fall to the right, much like an upside-down "U" shape. This visual guide is crucial as it sets the stage for accurate graph plotting.
Polynomial Zeros
Finding the zeros of a polynomial is another critical step. Zeros, also known as roots or solutions, are where the graph crosses or touches the x-axis. For the function \( g(x) = -x^2 + 10x - 16 \), we locate these points by setting the equation to zero. We get:\[0 = -x^2 + 10x - 16\]This can be factored into:\[0 = -(x - 2)(x - 8)\]The solutions are \( x = 2 \) and \( x = 8 \). These x-values are the zeros of the polynomial, meaning they are points on the graph where \( g(x) = 0 \). That means the graph will cross the x-axis at (2,0) and (8,0).
Knowing the zeros gives us anchor points for the graph and helps ensure our sketch will accurately reflect these critical intersections with the x-axis.
Continuous Curve Sketching
After identifying the end behavior via the Leading Coefficient Test and locating the zeros, it's time to bring the graph to life by sketching a continuous curve. This involves plotting several solution points to guide our drawing.
Start with the zeros, \( x = 2 \) and \( x = 8 \). Then select additional points around these zeros. Here, you can plot points such as \( x = 1 \), \( x = 5 \), and \( x = 9 \), and find their corresponding \( y \)-values by substituting back into the function \( g(x) \). These points will plot where the graph hits a particular \( y \)-value for the given \( x \).
With these points sketched, use the information about the falling ends from the Leading Coefficient Test to draw the curve through them in a smooth, continuous line. Remember, the graph should pass through each of these points and reflect the "upside-down U" shape you expect from its end behavior. Continuous curve sketching not only gives you a rough sketch but ensures your graph is reflecting the polynomial's characteristics accurately.

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Most popular questions from this chapter

In Exercises 51 - 54, write the polynomial (a) as the product of factors that are irreducible over the rationals, (b) as the product of linear and quadratic factors that are irreducible over the reals, and (c) in completely factored form. \( f(x) = x^4 - 3x^3 - x^2 - 12x - 20 \) (Hint: One factor is \( x^2 + 4 \).)

(a) Graph \( y = ax^2 \) for \( a = -2, -1, -0.5, 0.5, 1 \) and \( 2 \). How does changing the value of affect the graph? (b) Graph \( y = (x - h)^2 \) for \( h= -4, -2, 2, \) and \( 4 \). How does changing the value of \( h \) affect the graph? (c) Graph \( y = x^2 + k \) for \( k = -4, -2, 2, \) and \( 4 \). How does changing the value of \( k \) affect the graph?

In Exercises 35- 50, (a) find all the real zeros of the polynomial function, (b) determine the multiplicity of each zero and the number of turning points of the graph of the function, and (c) use a graphing utility to graph the function and verify your answers. \( f(x) = 2x^4 - 2x^2 - 40 \)

In Exercises 55 - 58, use the Remainder Theorem and synthetic division to find each function value. Verify your answers using another method. \( f(x) = 2x^3 - 7x + 3 \) (a) \( f(1) \) (b) \( f(-2) \) (c) \( f\left(\frac{1}{2}\right) \) (d) \( f(2) \)

In Exercises 75 - 80, (a) use the zero or root feature of a graphing utility to approximate the zeros of the function accurate to three decimal places,(b) determine one of the exact zeros, and (c) use synthetic division to verify your result from part (b), and then factor the polynomial completely. \( h(t) = t^3 - 2t^2 - 7t + 2 \)

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