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Chapter 2: Problem 64

Find a parabola with equation \(y=a x^{2}+b x+c\) that has slope 4 at \(x=1,\) slope \(-8\) at \(x=-1,\) and passes through the point \((2,15) .\)

Short Answer

Expert verified
The equation of the parabola is \( y = 3x^2 - 2x + 7 \).

Step by step solution

01

Derive the Derivative

The slope of the parabola is given by the derivative of the function. Start with the function \( y = ax^2 + bx + c \) and differentiate it with respect to \( x \). This gives \( \frac{dy}{dx} = 2ax + b \). The derivative represents the slope of the parabola at any point \( x \).
02

Use Slope Condition at x=1

From the problem, the slope at \( x = 1 \) is 4. Substitute \( x = 1 \) into the derivative and set it equal to 4: \( 2a(1) + b = 4 \). This simplifies to \( 2a + b = 4 \).
03

Use Slope Condition at x=-1

The slope at \( x = -1 \) is -8. Substitute \( x = -1 \) into the derivative and set it equal to -8: \( 2a(-1) + b = -8 \). This simplifies to \( -2a + b = -8 \).
04

Solve the System of Equations for a and b

Now solve the system of equations from Step 2 and Step 3:\[\begin{align*}2a + b &= 4 \-2a + b &= -8\end{align*}\]Subtract the second equation from the first:\[ 4a = 12 \]Thus, \( a = 3 \). Substitute \( a = 3 \) into \( 2a + b = 4 \):\[ 2(3) + b = 4 \quad \Rightarrow \quad b = -2 \].
05

Use the Point Condition to Find c

The parabola passes through the point \((2, 15)\). Substitute \( x = 2 \), \( y = 15 \), \( a = 3 \), and \( b = -2 \) into the equation: \( 15 = 3(2)^2 - 2(2) + c \). Simplify to find \( c \): \( 15 = 12 - 4 + c \) which gives \( c = 7 \).
06

Formulate the Equation of the Parabola

Substitute \( a = 3 \), \( b = -2 \), and \( c = 7 \) into the general equation for the parabola: \( y = 3x^2 - 2x + 7 \).

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Key Concepts

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

Derivative
The derivative of a function is a crucial mathematical tool that helps us find the slope or rate of change of that function at any given point. In the context of a parabola, the derivative provides insights into how the shape of the curve changes as we move along the x-axis. For the parabola equation given by \[ y = ax^2 + bx + c, \]its derivative is expressed as\[ \frac{dy}{dx} = 2ax + b. \]This equation tells us the slope of the tangent line to the parabola at any point \( x \). By understanding and utilizing the derivative, we can solve various problems involving parabolas with ease.
Slope Condition
The slope condition involves determining the slope of a function at specific points. In this exercise, the parabola must have a slope of 4 at \( x = 1 \) and a slope of -8 at \( x = -1 \). The derivative allows us to set conditions for these slopes. For instance:- At \( x = 1 \), the slope is given by substituting into the derivative, resulting in \[ 2a(1) + b = 4, \] which simplifies to \( 2a + b = 4 \).- At \( x = -1 \), the slope \[ 2a(-1) + b = -8, \] simplifies to \( -2a + b = -8 \).These slope conditions allow us to create equations that describe the behavior of the parabola's curve at specific x-values, thus helping us identify the coefficients \( a \) and \( b \).
System of Equations
Once we have our slope conditions, we can set up a system of equations to solve for the parabola's coefficients. The equations derived from the slope conditions are:\[ \begin{align*}2a + b &= 4 \-2a + b &= -8\end{align*} \]To solve these, you can use methods like substitution or elimination. In this instance, subtracting the second equation from the first simplifies directly to:- \( 4a = 12, \)which solves to give \( a = 3 \). Substituting \( a = 3 \) back into the first equation, we find:- \( 2(3) + b = 4, \)so \( b = -2 \). These solutions fit neatly into our equation for the parabola.
Point Condition
The point condition requires the parabola to pass through a specific point, given in this exercise as \((2, 15)\). We now need to substitute this point and earlier found coefficients (\( a = 3 \), \( b = -2 \)) into the parabola equation to find \( c \):\[ 15 = 3(2)^2 - 2(2) + c. \]Simplifying this gives- \( 15 = 12 - 4 + c \)from which we find that \( c = 7 \).Putting it all together, the final equation of the parabola becomes:\[ y = 3x^2 - 2x + 7. \]This encapsulates all given conditions: the slopes at \( x=1 \) and \( x=-1 \), and the pass-through point \((2, 15)\).

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Most popular questions from this chapter

A faucet is filling a hemispherical basin of diameter 60 \(\mathrm{cm}\) with water at a rate of 2 \(\mathrm{L} / \mathrm{min}\) . Find the rate at which the water is rising in the basin when it is half full. [Use the following facts: 1 \(\mathrm{L}\) is 1000 \(\mathrm{cm}^{3} .\) The volume of the portion of a sphere with radius \(r\) from the bottom to a height \(h\) is \(V=\pi\left(r h^{2}-\frac{1}{3} h^{3}\right),\) as we will show in Chapter \(7 . ]\)

Some of the highest tides in the world occur in the Bay of Fundy on the Atlantic Coast of Canada. At Hopewell Cape the water depth at low tide is about 2.0 \(\mathrm{m}\) and at high tide it is about 12.0 \(\mathrm{m} .\) The natural period of oscillation is a little more than 12 hours and on June \(30,2009,\) high tide occurred at \(6 : 45\) AM. This helps explain the following model for the water depth \(D\) (in meters) as a function of the time \(t\) (in hours after midnight) on that day: $$D(t)=7+5 \cos [0.503(t-6.75)]$$ How fast was the tide rising (or falling at the following times? $$\begin{array}{ll}{\text { (a) } 3 : 00 \mathrm{AM}} & {\text { (b) } 6 : 00 \mathrm{AM}} \\ {\text { (c) } 9 : 00 \mathrm{AM}} & {\text { (d) Noon }}\end{array}$$

$$ \begin{array}{l}{\text { If } g \text { is a twice differentiable function and } f(x)=x g\left(x^{2}\right),} \\ {\text { find } f^{\prime \prime} \text { in terms of } g, g^{\prime}, \text { and } g^{\prime \prime}}\end{array} $$

(a) Find equations of both lines through the point \((2,-3)\) that are tangent to the parabola \(y=x^{2}+x\) . (b) Show that there is no line through the point \((2,7)\) that is tangent to the parabola. Then draw a diagram to see why.

A kite 100 \(\mathrm{ft}\) above the ground moves horizontally at a speed of 8 \(\mathrm{ft} / \mathrm{s}\) . At what rate is the angle between the string and the horizontal decreasing when 200 \(\mathrm{ft}\) of string has been let out?
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