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Chapter 2: Problem 8

Given that \(f(-2)=3\) and \(f^{\prime}(-2)=-4,\) find an equation for the tangent line to the graph of \(y=f(x)\) at \(x=-2\)

Short Answer

Expert verified
The equation of the tangent line is \(y = -4x - 5\).

Step by step solution

01

Understand the Point-Slope Form

The equation of a line in point-slope form is given by \[ y - y_1 = m(x - x_1) \]where \((x_1, y_1)\) is a point on the line, and \(m\) is the slope of the line.
02

Identify the Point on the Graph

Since we know that when \(x = -2\), \(f(-2) = 3\), the point \((-2, 3)\) lies on the graph of the function \(y = f(x)\). This point will be \((x_1, y_1)\) in the point-slope form.
03

Identify the Slope of the Tangent Line

The derivative \(f'(x)\) at a point gives the slope of the tangent line at that point. We are given that \(f'(-2) = -4\), so the slope \(m\) of the tangent line is \(-4\).
04

Plug Values into Point-Slope Form

Substitute \((x_1, y_1) = (-2, 3)\) and \(m = -4\) into the point-slope form equation: \[ y - 3 = -4(x + 2) \]
05

Simplify the Equation

Distribute and simplify the equation to put it in slope-intercept form \(y = mx + b\):\[ y - 3 = -4x - 8 \]Add 3 to both sides:\[ y = -4x - 5 \]

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Key Concepts

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

Point-Slope Form
The point-slope form of a linear equation is a powerful tool in geometry and calculus. It's particularly useful for finding the equation of a line when you know one point on the line and the slope. The point-slope form is written as:\[ y - y_1 = m(x - x_1) \]Here, \((x_1, y_1)\) is your known point, and \(m\) is the slope of the line.
  • The left side of the equation, \(y - y_1\), signifies the vertical distance from your point to any point \((x, y)\) on the line.
  • The right side, \(m(x - x_1)\), represents the horizontal movement multiplied by the slope.
This form directly shows how every change in \(x\) affects \(y\), making it easy to see how altering variables affects the line's direction. By substituting known values, you can derive a complete equation of the line.
Derivative
In calculus, the derivative is a critical concept representing the rate of change or slope of a curve at any specific point. When you have a function like \(f(x)\), the derivative, denoted as \(f'(x)\), shows how \(f(x)\) changes as \(x\) changes.
  • The derivative is fundamental in finding the tangent line, which touches the curve at precisely one point without crossing it.
  • It tells us the slope at this very point, allowing us to draw a precise line that can model linear approximations of curves.
For example, if you know \(f'(-2) = -4\), it implies that at \(x = -2\), the slope of the tangent to the curve is \(-4\). This slope is then used in conjunction with a point on the curve to determine the tangent line equation using point-slope form.
Slope-Intercept Form
The slope-intercept form of a linear equation is perhaps the most familiar form of a line equation to many students. It is expressed as:\[ y = mx + b \]This form directly reveals two critical pieces of information: the slope \(m\) and the y-intercept \(b\).
  • The slope \(m\) informs you how steep the line is and in which direction it tilts.
  • The y-intercept \(b\) provides the exact starting point where the line crosses the y-axis.
After determining a line equation in point-slope form, it's often insightful or required to convert it into slope-intercept form to visualize or further analyze the line. This involves distributing and simplifying the point-slope expression to align with the \(y = mx + b\) format. Doing so provides a clearer, overall picture of the line's behavior within the coordinate plane.

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Most popular questions from this chapter

Find \(d y / d x\) $$ y=\frac{\sin x}{\sec (3 x+1)} $$

Find \(f^{\prime}(x)\) $$ f(x)=\sin \left(\frac{1}{x^{2}}\right) $$

Find all points where \(f\) fails to be differentiable. Justify your answer. \(\begin{array}{llll}{\text { (a) } f(x)=|3 x-2|} & {\text { (b) } f(x)=\left|x^{2}-4\right|} & {}\end{array}\)

(a) Use a graphing utility to obtain the graph of the function \(f(x)=\sin x^{2} \cos x\) over the interval \([-\pi / 2, \pi / 2] .\) (b) Use the graph in part (a) to make a rough sketch of the graph of \(f^{\prime}\) over the interval. (c) Find \(f^{\prime}(x),\) and then check your work in part (b) by using the graphing utility to obtain the graph of \(f^{\prime}\) over the interval. (d) Find the equation of the tangent line to the graph of \(f\) at \(x=1,\) and graph \(f\) and the tangent line together over the interval.

Find the indicated derivative. $$ x=\csc ^{2}\left(\frac{\pi}{3}-y\right) ; \text { find } \frac{d x}{d y} $$
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