Problem 20 - Calculus: Early Transcendentals

Question

Find an equation of the tangent line to the graph of \( y = g(x) \) at \( x = 5 \) if \( g(5) = -3 \) and \( g'(5) = 4 \).

Step-by-Step Solution

Answer

The equation of the tangent line is \(y = 4x - 23\).

Step 1: Identify the Point

We begin by identifying the point at which the tangent line touches the graph. The point is given by the coordinates \((x, y)\) where \(x = 5\) and we are given that \(y = g(5) = -3\). Hence, the point is \((5, -3)\).

Step 2: Determine the Slope

The slope of the tangent line is given by the derivative of the function at the point of tangency. We are given that \(g'(5) = 4\). Therefore, the slope \(m\) of the tangent line is \(4\).

Step 3: Use the Point-Slope Formula

Having the slope \(m = 4\) and a point \((5, -3)\), we use the point-slope form of a line equation, which is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is our point. Plug in the values: \(y - (-3) = 4(x - 5)\).

Step 4: Simplify the Equation

We simplify the equation from the point-slope form. Starting with: \(y + 3 = 4(x - 5)\), distribute the \(4\): \(y + 3 = 4x - 20\). Subtract \(3\) from both sides to get the equation in slope-intercept form: \(y = 4x - 23\).

Key Concepts

Derivative

The derivative is a fundamental concept in calculus that represents the rate of change or the slope of a function at any given point. When you think about drawing a tangent line at a specific point on the graph of a function, you're really interested in how steep the graph is at that point.

The derivative of a function at a particular point tells you exactly this steepness or inclination. For example, given the function \( y = g(x) \), the expression \( g'(x)\) embodies the derivative of \(g\) with respect to \(x\).

If you know \( g'(5) = 4 \), it means that at \( x = 5 \), the rate of change of the function is 4. This is the slope of the tangent line at this point. In essence, a derivative provides a snapshot of the function's behavior at a precise location, helping us understand how the function moves.

Point-Slope Form

The point-slope form is a standard way of expressing the equation of a line. This form makes it easy to write the equation when you know a point on the line and its slope.

The formula for the point-slope form is:

• \( y - y_1 = m(x - x_1) \)

Here, \((x_1, y_1)\) represents the coordinate of the point on the line, and \( m \) is the slope.

In our exercise, the point we have is \((5, -3)\) and the slope \( m \) is 4. By plugging in these values, we can easily find the equation of the tangent line which touches the graph exactly at this point. The initial equation \( y - (-3) = 4(x - 5) \) is simple to set up using the known values. After setting it up, a little algebraic manipulation will lead to the slope-intercept form.

Slope-Intercept Form

The slope-intercept form is probably the easiest and most widely used expression of a line. This is because it clearly shows both the slope of the line and where it crosses the y-axis.

The formula is:

• \( y = mx + b \)

Here's how it works:

• \( m \) is the slope of the line.
• \( b \) is the y-intercept, or where the line crosses the y-axis.

In our step-by-step solution, we rearranged the equation from point-slope form to slope-intercept form, ending up with \( y = 4x - 23 \). Here, 4 is the slope, telling us that for every unit increase in \( x \), \( y \) increases by 4.

Meanwhile, \( -23 \) is the y-intercept, meaning the line crosses the y-axis at -23. So, this form of the equation gives us a clear and straightforward picture of the line's behavior and position.