Question

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

Short answer

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

Step-by-step solution

Determine the derivative

To find the equation of the tangent line, we first need the derivative of the function. The function given is \(y = x^{2} - 2x + 2\). We find the derivative using basic differentiation rules: \(y' = 2x - 2\). This derivative gives us the slope \(m\) of the tangent line at any point \(x\) on the curve.

Evaluate the derivative at the given point

Substitute \(x = 1\) into the derivative to find the slope of the tangent line at that point. \(y' = 2(1) - 2 = 0\). Thus, the slope of the tangent line at \((1, 1)\) is \(0\).

Write the tangent line equation

The general equation of a line is \(y = mx + c\), where \(m\) is the slope and \(c\) is the y-intercept. With the point \((1, 1)\) and the slope \(m = 0\), the line equation is \(y = 0 \cdot x + c\). Substituting the point \((1, 1)\) into this equation, we find \(1 = c\). Therefore, the equation of the tangent line is \(y = 1\), which is a horizontal line passing through \((1, 1)\).

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus. It helps us understand how a function changes at any given point. The process of differentiation involves finding the derivative of a function. A derivative gives the rate at which one quantity changes with respect to another. For functions like polynomials, using basic rules can easily find the derivative.

• Power rule: If you have a function \(f(x) = x^n\), then its derivative is \(f'(x) = nx^{n-1}\).
• Sum rule: The derivative of a sum of functions is the sum of the derivatives.
• Constant rule: The derivative of a constant is always zero.

In our exercise, we differentiated \(y = x^{2} - 2x + 2\). Applying the rules, we find the derivative \(y' = 2x - 2\). This expression tells us how the function \(y\) changes as \(x\) changes.

Slope of the Tangent

The slope of the tangent line at a particular point on a curve is crucial. It indicates how steep the curve is at that exact location. Mathematically, the slope at a point is the value of the derivative evaluated at that point. In our problem, the function \(y = x^{2} - 2x + 2\) was differentiated to get \(y' = 2x - 2\).
To find the slope of the tangent line at the point \((1,1)\), we plug \(x = 1\) into the derivative:

• \(y' = 2(1) - 2 = 0\).

This result, \(0\), tells us that the slope of the tangent at \((1,1)\) is zero, indicating a horizontal line. A zero slope suggests no vertical change in the line at that point.

Derivative Evaluation

Evaluating the derivative at a specific point allows us to determine important properties like the tangent slope. This process involves substituting the particular value of \(x\) into the derivative to see how the original function behaves.
In our example, we had \(y' = 2x - 2\) and checked it for \(x = 1\). We found the slope of the tangent was zero:

• \(y' = 2(1) - 2 = 0\).

This evaluation confirmed that the tangent at \((1, 1)\) is horizontal. Knowing how to evaluate derivatives quickly allows you to find tangent lines efficiently. It helps apply calculus concepts to solve real-world problems, like cutting-edge technologies and engineering solutions.