/*! 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 62 Let \(f(x)=-2 x^{2}+a x+b .\) De... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(f(x)=-2 x^{2}+a x+b .\) Determine the constants \(a\) and \(b\) such that \(f\) has a relative maximum at \(x=2\) and the relative maximum value is 4 .

Short Answer

Expert verified
The constants \(a\) and \(b\) are 10 and -8, respectively, resulting in the function \(f(x) = -2x^2 + 10x - 8\).

Step by step solution

01

Use the given x-coordinate of the vertex to determine the value of a.

We are given that the maximum occurs at \(x = 2\). Using the formula for the \(x\)-coordinate of the vertex, we have: \(2 = \frac{-b}{2(-2)}\) Now, solve for \(b\): \(2 = \frac{-b}{-4}\) Multiplying both sides by -4, we find that \(b = -8\).
02

Use the maximum value at the vertex to determine the value of a.

We are given that the relative maximum value is 4. Now that we know \(b = -8\), we can plug the x-coordinate of the vertex (2) into our equation and set the function equal to 4: \(4 = -2(2)^2 + a(2) - 8\) Solve for \(a\): \(4 = -8 + 2a - 8\) \(20 = 2a\) \(a = 10\)
03

Write the function with the determined values of a and b.

Now that we have found \(a = 10\) and \(b = -8\), we can write out the function: \(f(x) = -2x^2 + 10x - 8\) So, the function that satisfies the given conditions is \(f(x) = -2x^2 + 10x - 8\).

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Key Concepts

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

Vertex Form of a Quadratic
Understanding the vertex form of a quadratic function is essential for analyzing its graph and properties. A quadratic function is typically written in the form of \( ax^2 + bx + c \), known as the standard form. However, the vertex form, \( a(x-h)^2 + k \), provides insightful details such as the location of the vertex and the direction of opening (upward or downward) of the parabola.

In the vertex form, \( (h, k) \) represents the coordinates of the vertex of the parabola, which can be a maximum or a minimum point depending on the leading coefficient \( a \). If \( a < 0 \), the parabola opens downwards and the vertex is the maximum point; conversely, if \( a > 0 \), it opens upward and the vertex is the minimum point.

To convert from standard to vertex form, one can complete the square or use formulas for the vertex coordinates: \( h = -\frac{b}{2a} \) and \( k = f(h) \). Completing the square involves creating a perfect square trinomial from the quadratic terms in the standard form and then simplifying accordingly. This is a powerful technique in finding the vertex, especially for optimization problems.
Optimization of Quadratic Functions
Optimization is a mathematical strategy used to find the most efficient or least costly solution to a problem. In the context of quadratic functions, optimization often involves finding the maximum or minimum values of the function, which correspond to the vertex of its graph.

Since the graph of a quadratic function is a parabola, the optimization process focuses on determining the vertex, as this point will represent either the highest or lowest possible value (the optimum) depending on whether the parabola opens upwards or downwards. If the leading coefficient \( a \) of the quadratic function is negative, the function reaches its maximum value at the vertex; if \( a \) is positive, the function reaches its minimum.

To optimize a quadratic function given in standard form, you can use the vertex formula or convert to vertex form to identify the optimal value easily. Doing so enables you to solve various practical problems involving area, profit, trajectory, and even pricing models in economics.
Solving Quadratic Equations
Solving quadratic equations is a foundational skill in algebra that involves finding the values of the variable \( x \) that satisfy the equation \( ax^2 + bx + c = 0 \). There are several methods to solve quadratic equations: factoring, using the quadratic formula, completing the square, or graphically analyzing the function.

Factoring is often the first method attempted as it involves finding two binomials that multiply to give the quadratic equation. When factoring is not straightforward or possible, the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), can be utilized. This formula provides the exact solutions for any quadratic equation and is especially useful when the equation does not factorize neatly.

Another method, completing the square, transforms the standard form of the quadratic into the vertex form. This not only helps find the solutions by isolating \( x \) but also provides clarity on the properties of the quadratic function, which is useful in the analysis of the function's behavior.
Applications of Derivatives
In calculus, derivatives represent how a function changes with respect to a variable, and they are crucial in analyzing the behavior of functions. The applications of derivatives are vast and include finding slopes of curves, determining rates of change, and solving optimization problems.

For quadratic functions, the derivative is particularly useful in finding the maximum or minimum values which are points where the slope of the tangent to the graph is zero. By taking the derivative of the quadratic function and setting it equal to zero, you can solve for \( x \) to find the critical points. In the context of the given example, derivatives help to confirm that the vertex of the quadratic function is indeed the point of relative maximum or minimum.

Moreover, the second derivative can be employed to determine the concavity of a function, which indicates whether the critical point found is a maximum or minimum. If the second derivative is negative at that point, the function is concave down, indicating a relative maximum. If positive, it is concave up, indicating a relative minimum. Derivatives are thus a tool of immense power in various scientific and engineering fields, as well as in economics and optimization problems.

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Most popular questions from this chapter

Mass of a Moving Particle The mass \(m\) of a particle moving at a speed \(v\) is related to its rest mass \(m_{0}\) by the equation $$ m=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}} $$ where \(c\), a constant, is the speed of light. Show that $$ \lim _{v \rightarrow c^{-}} \frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\infty $$ thus proving that the line \(v=c\) is a vertical asymptote of the graph of \(m\) versus \(v .\) Make a sketch of the graph of \(m\) as a function of \(v\).

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Chemical Mixtures Two chemicals react to form another chemical. Suppose that the amount of the chemical formed in time \(t\) (in hours) is given by $$ x(t)=\frac{15\left[1-\left(\frac{2}{3}\right)^{3 t}\right]}{1-\frac{1}{4}\left(\frac{2}{3}\right)^{3 t}} $$ where \(x(t)\) is measured in pounds. a. Plot the graph of \(x\) using the viewing window \([0,10] \times[0,16] .\) b. Find the rate at which the chemical is formed when \(t=1\). c. How many pounds of the chemical are formed eventually?

Use the appropriate precise definition to prove the statement. $$ \lim _{x \rightarrow \infty} \frac{x}{x^{2}+1}=0 $$

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