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Chapter 4: Problem 66

Determine, without graphing, whether the given quadratic function has a maximum value or a minimum value, and then find the value. \(f(x)=-2 x^{2}+12 x\)

Short Answer

Expert verified
The function has a maximum value of 18.

Step by step solution

01

Determine the form of the quadratic function

The given quadratic function is \(f(x)=-2 x^{2}+12 x\). It is in the standard form \(f(x)=ax^2 + bx + c\).Here, \(a = -2\), \(b = 12\), and \(c = 0\).
02

Identify the coefficient of the quadratic term

Look at the coefficient of the quadratic term, which is \(a=-2\). Since \(a\) is negative, the parabola opens downwards. This means the function has a maximum value.
03

Find the vertex of the quadratic function

The vertex of a quadratic function \(f(x) = ax^2 + bx + c\) can be found at \(x = -\frac{b}{2a}\). Plug \(a = -2\) and \(b = 12\) into the formula:\[ x = -\frac{12}{2(-2)} = \frac{12}{4} = 3 \]
04

Calculate the value of the function at the vertex

Substitute \(x = 3\) back into the function to find the maximum value:\[ f(3) = -2(3)^2 + 12(3) = -2(9) + 36 = -18 + 36 = 18 \]So, the maximum value of the function is 18.

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Key Concepts

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

vertex formula
In a quadratic function, the vertex is a crucial point where the function reaches its maximum or minimum value. The vertex formula helps us find the x-coordinate of this vertex. For any quadratic function in the standard form: \( f(x) = ax^2 + bx + c \), you can find the x-coordinate of the vertex using the formula:
This formula plays a significant role in analyzing quadratic functions. Plugging the constants from the quadratic equation into this formula directly provides us with the vertex x-coordinate. Knowing this, you can easily determine the function's value at this point to find extremums.
maximum value
In the given quadratic function \( f(x) = -2x^2 + 12x \), we first noted that the coefficient of \(x^2\) is negative (\(a = -2\)). If the leading coefficient \(a\) is negative, the parabola opens downwards, indicating a maximum value at the vertex.
After determining the parabola opens downwards, we've calculated the x-coordinate of the vertex using the vertex formula. By substituting this x-coordinate back into the original function, we determine the maximum function value.
We substituted \(x = 3\) back into the function to get \( f(3) = -2(3)^2 + 12(3) = 18 \), giving us the maximum value of \(18\).
parabola
A quadratic function graph is called a parabola. For the given function \( f(x) = -2x^2 + 12x \), the parabola opens downwards because the coefficient of \(x^2\) (which is \(a\)) is negative. This influences how we determine the vertex and the extremums of the function.
Parabolas have some unique properties:
  • The direction of opening (upwards or downwards) depends on the sign of \(a\).
  • The vertex is the highest or lowest point of the parabola.
  • The axis of symmetry of a parabola is a vertical line that passes through the vertex.

Understanding how the parabola's direction and the vertex's position affect the quadratic function is important. It helps to find the maximum or minimum values without graphing. In summary, recognizing these properties aids in solving and graphing quadratic equations effectively and efficiently.

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Most popular questions from this chapter

The marginal cost of a product can be thought of as the cost of producing one additional unit of output. For example, if the marginal cost of producing the \(50 t h\) product is \(\$ 6.20,\) it costs \(\$ 6.20\) to increase production from 49 to 50 units of output. Suppose the marginal cost \(C\) (in dollars) to produce \(x\) thousand digital music players is given by the function $$ C(x)=x^{2}-140 x+7400 $$ (a) How many players should be produced to minimize the marginal cost? (b) What is the minimum marginal cost?

Revenue The John Deere company has found that the revenue from sales of heavy- duty tractors is a function of the unit price \(p,\) in dollars, that it charges. The revenue \(R,\) in dollars, is given by $$ R(p)=-\frac{1}{2} p^{2}+1900 p $$ (a) At what prices \(p\) is revenue zero? (b) For what range of prices will revenue exceed \(\$ 1,200,000 ?\)

(a) Graph fand g on the same Cartesian plane. (b) Solve \(f(x)=g(x)\) (c) Use the result of part (b) to label the points of intersection of the graphs of fand \(g\). (d) Shade the region for which \(f(x)>g(x)\); that is, the region below fand above \(g\). \(f(x)=-x^{2}+4 ; \quad g(x)=-2 x+1\)

The graph of the function \(f(x)=a x^{2}+b x+c\) has vertex at (1,4) and passes through the point \((-1,-8) .\) Find \(a, b\), and \(c\)

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Use a graphing utility to graph \(f(x)=x^{3}-8 x^{2}+13 x-2\) over the interval \([-2,8] .\) Then, approximate any local maximum values and local minimum values, and determine where \(f\) is increasing and where \(f\) is decreasing. Round answers to two decimal places.
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