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Chapter 3: Problem 11

Find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. $$f(x)=x^{2}+1, \quad(2,5)$$

Short Answer

Expert verified
The slope is 4, and the tangent line equation is \(y = 4x - 3\).

Step by step solution

01

Understand the Problem

We are given the function \(f(x) = x^2 + 1\) and need to find the slope of the graph at the point \((2, 5)\). Next, we need to find an equation for the tangent line at this point.
02

Find the Derivative

To find the slope of the tangent line, we first need the derivative of the function \(f(x)\). The derivative, \(f'(x)\), represents the slope of the tangent line at any point \(x\). For \(f(x) = x^2 + 1\), the derivative is \(f'(x) = 2x\).
03

Evaluate the Derivative at the Given Point

We evaluate \(f'(x)\) at \(x = 2\) to find the slope of the tangent line at the point \((2, 5)\). Substituting, we get \(f'(2) = 2(2) = 4\). So, the slope of the tangent line at \((2, 5)\) is 4.
04

Use Point-Slope Form to Write the Tangent Line Equation

The point-slope form of a line is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line, and \(m\) is the slope. Here, \((x_1, y_1) = (2, 5)\) and \(m = 4\). Substituting these values gives us \(y - 5 = 4(x - 2)\).
05

Simplify the Tangent Line Equation

Solve the equation \(y - 5 = 4(x - 2)\) to express it in the slope-intercept form \(y = mx + b\). Expanding gives \(y - 5 = 4x - 8\), and then \(y = 4x - 3\). Thus, the equation of the tangent line is \(y = 4x - 3\).

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Key Concepts

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

Slope
The concept of slope is a fundamental element when we talk about tangent lines. Essentially, the slope is a measure of how steep a line is. In mathematical terms, it's the ratio of the vertical change (rise) to the horizontal change (run) between two points on a line.

For linear equations, the slope is constant, but when dealing with curves, like the function given, the slope varies. This is where the tangent line comes in.

The slope of the tangent line to a curve at a particular point is the steepness of the curve precisely at that point. It helps identify how quickly the function is changing at that point.

In our exercise, we found the slope at the point \(2, 5\) on the curve \(f(x) = x^2 + 1\) to be 4. This illustrates the curve's rate of change at that specific location on the graph.
Derivative
Derivatives are a crucial concept in calculus and they help us find the slope of a curve. Think of the derivative as a tool that finds the rate at which a function's value is changing at any point.

To put it simply, if you have a function, like \(f(x) = x^2 + 1\), the derivative of that function tells you the slope of the tangent line at any given x-value. In this case, the derivative of \(f(x)\) was calculated to be \(f'(x) = 2x\).

This derivative function can be used to find the specific slope at any point along the curve. For instance, to find the slope at \(x = 2\), we substitute this x-value into the derivative to get \(f'(2) = 4\). This result shows that at the point \(2, 5\), the graph has a slope of 4.
Point-Slope Form
The point-slope form is a helpful formula for writing the equation of a line when you know the slope and a single point on the line. This form is represented as \(y - y_1 = m(x - x_1)\), where \(m\) is the slope, and \( (x_1, y_1) \) is a known point.

In our problem, after determining that the slope of the tangent line at the given point \(2, 5\) is 4, we used point-slope form to write the line's equation.

By substituting \(m = 4\), \(x_1 = 2\), and \(y_1 = 5\) into the formula, we obtained \(y - 5 = 4(x - 2)\).

This formula can be rearranged into a more familiar slope-intercept form \(y = mx + b\), which becomes \(y = 4x - 3\) after simplification. The point-slope form makes it straightforward to quickly find equations of lines, especially those tangent to curves.

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