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Chapter 2: Problem 49

Find the equation of the tangent line to \(y=x^{2}-2 x+2\) at the point (1,1) .

Short Answer

Expert verified
The equation of the tangent line is \(y = 1\).

Step by step solution

01

Find the Derivative

To find the equation of the tangent line, we need to calculate the derivative of the function \(y = x^2 - 2x + 2\). The derivative is \(y' = \frac{d}{dx}(x^2 - 2x + 2) = 2x - 2\).
02

Evaluate the Derivative at the Point

Substitute \(x = 1\) into the derivative to find the slope of the tangent line at the point (1,1). Calculate \(m = 2(1) - 2 = 0\). So, the slope of the tangent line is 0.
03

Use the Point-Slope Formula

The point-slope formula for a line is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the point and \(m\) is the slope. Substitute \((x_1, y_1) = (1, 1)\) and \(m = 0\):\[y - 1 = 0(x - 1)\].
04

Simplify the Equation

Simplify the equation from Step 3:\[y - 1 = 0\] which simplifies to \[y = 1\]. This is the equation of the tangent line.

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Key Concepts

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

Derivatives
Derivatives are fundamental in calculus. They represent the rate at which a function is changing at any given point and are denoted by the symbol \( f'(x) \) or \( \frac{df}{dx} \).
In simpler terms, if you think of a function as describing the path of a moving car, the derivative tells you the car's speed at each moment. This concept is vital for understanding how different phenomena evolve over time.
  • To find a derivative, you generally apply rules such as the power rule, product rule, or chain rule.
  • In this exercise, the power rule is used to derive \( y = x^2 - 2x + 2 \). The power rule states that the derivative of \( x^n \) is \( nx^{n-1} \).
So, applying the power rule:
  • Derivative of \( x^2 \) is \( 2x \).
  • Derivative of \( -2x \) is \( -2 \).
  • Derivative of a constant \( +2 \) is \( 0 \).
Therefore, the derivative of the function is \( y' = 2x - 2 \). This expression helps in finding the slope of the tangent line at any point \( x \) on the curve.
Tangent Line
A tangent line is a straight line that just "touches" a curve at a given point. It mirrors the slope of the curve precisely at that point, making it a powerful tool in calculus for approximation and analysis.
When dealing with differentiable functions, the slope of the tangent line at a given point is the value of the derivative of the function at that point.
To find the tangent line in our exercise:
  • We calculate the derivative to determine the slope formula: \( y' = 2x - 2 \).
  • We then substitute \( x = 1 \) into this formula, since that's our point of interest, yielding \( m = 0 \), which indicates a horizontal tangent line at point \((1,1)\).
Since the slope is 0, it means the tangent line does not "rise" or "fall" but is flat, parallel to the x-axis.
Visualizing this helps students see how the curve behaves locally around the point where it meets the tangent line.
Point-Slope Formula
The point-slope formula is a staple in linear equations. It's used to construct the equation of a line when given a point and the slope. The formula is \( y - y_1 = m(x - x_1) \).
This formula relies on the known point \( (x_1, y_1) \) and the slope \( m \) to find the line's equation.
To solve the exercise:
  • We have the point \( (1, 1) \).
  • The slope \( m = 0 \), as calculated from the derivative.
Substituting these into the point-slope formula yields:
  • \( y - 1 = 0(x - 1) \) which simplifies to \( y = 1 \).
This means that the tangent line is a horizontal line crossing the y-axis at 1, succinctly represented by the equation \( y = 1 \).
The point-slope formula efficiently transforms the mathematical slope information into a readable line equation that describes the behavior of the tangent at the chosen point.

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Most popular questions from this chapter

Find the equation of the tangent line to \(y=\cos x\) at \(x=1\)

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