If h(1) = 2 and h′(1) = 3, find an equation of the line tangent to the graph of h at x = 1.

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Step 1: Recall that the equation of a line in point-slope form is given by \( y - y_1 = m(x - x_1) \), where \( m \) is the slope of the line and \( (x_1, y_1) \) is a point on the line.

Step 2: Identify the point \((x_1, y_1)\) on the tangent line. Since \( h(1) = 2 \), the point is \((1, 2)\).

Step 3: Determine the slope \( m \) of the tangent line. The slope of the tangent line at \( x = 1 \) is given by the derivative \( h'(1) \), which is 3.

Step 4: Substitute the point \((1, 2)\) and the slope \( m = 3 \) into the point-slope form equation: \( y - 2 = 3(x - 1) \).

Step 5: Simplify the equation if needed to express it in a different form, such as slope-intercept form \( y = mx + b \), by distributing and rearranging terms.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Tangent Line

A tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. The slope of the tangent line is equal to the derivative of the function at that point. In this case, the tangent line to the graph of h at x = 1 will have a slope equal to h′(1).

Derivative

The derivative of a function at a specific point measures the rate at which the function's value changes as its input changes. It is denoted as h′(x) and provides the slope of the tangent line at any point on the graph of the function. For this problem, h′(1) = 3 indicates that the slope of the tangent line at x = 1 is 3.

Point-Slope Form

The point-slope form of a linear equation is used to express the equation of a line given a point on the line and its slope. It is written as y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope. In this case, using the point (1, 2) and the slope 3, we can derive the equation of the tangent line.

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