Problem 17 - Precalculus: Mathematics for Calculus

Question

Find an equation of the tangent line to the curve at the given point. Graph the curve and the tangent line. $$y=\sqrt{x+3}, \quad \text { at }(1,2)$$

Step-by-Step Solution

Answer

The equation of the tangent line is \( y = \frac{1}{4}x + \frac{7}{4} \).

Step 1: Understand the Problem

We need to find the tangent line to the curve given by the function \( y = \sqrt{x+3} \) at the point \((1, 2)\). The tangent line will intersect the curve at this point and have the same slope as the curve at this point.

Step 2: Find the Derivative of the Function

To determine the slope of the tangent line, we first need to find the derivative of the function \( y = \sqrt{x+3} \). Recall that the derivative \( y' \) of \( y = (x+3)^{1/2} \) is found using the power rule: \( y' = \frac{1}{2}(x+3)^{-1/2} \). Thus, the derivative is \( y' = \frac{1}{2\sqrt{x+3}} \).

Step 3: Evaluate the Derivative at the Given Point

Next, we evaluate the derivative at \( x = 1 \) to find the slope of the tangent line at the point \((1, 2)\). Substitute \( x = 1 \) into the derivative: \( y'(1) = \frac{1}{2\sqrt{1+3}} = \frac{1}{2\cdot 2} = \frac{1}{4} \). Therefore, the slope of the tangent line is \( \frac{1}{4} \).

Step 4: Use the Point-Slope Form to Find the Equation of the Tangent Line

With the slope \( m = \frac{1}{4} \) and the point \((1, 2)\), we use the point-slope form of a line, which is \( y - y_1 = m(x - x_1) \). Substitute \( m = \frac{1}{4} \), \( x_1 = 1 \), and \( y_1 = 2 \) into the equation: \( y - 2 = \frac{1}{4}(x - 1) \). Simplify to get \( y = \frac{1}{4}x + \frac{7}{4} \).

Step 5: Graph the Curve and the Tangent Line

Finally, we plot the curve \( y = \sqrt{x+3} \) and the tangent line \( y = \frac{1}{4}x + \frac{7}{4} \). The curve is a horizontal stretch of the basic square root function that passes through the point \((1,2)\), while the tangent line crosses the curve at this point with a slope of \( \frac{1}{4} \). Ensure both are accurately graphed for visual confirmation of the solution.

Key Concepts

Derivative

The derivative of a function essentially tells us the slope of the function at any given point. It's a fundamental tool in calculus. Knowing the derivative helps us understand how the function changes as it progresses along the graph. In the exercise involving the function \( y = \sqrt{x+3} \), finding the derivative unveils the slope of the tangent line at the point \((1, 2)\).

To find the derivative of \( y = \sqrt{x+3} \), we use the power rule. Consider \( y = (x+3)^{1/2} \). Applying the power rule for derivatives, \( y' = \frac{d}{dx}(x+3)^{1/2} \), results in:

• Bring down the exponent: \( \frac{1}{2} \).
• Reduce the exponent by one: \((x+3)^{-1/2} \).
• Combine these results: \( y' = \frac{1}{2}(x+3)^{-1/2} \).

Thus, the derivative becomes \( y' = \frac{1}{2\sqrt{x+3}} \). Evaluating this at \( x = 1 \) gives us the slope of the tangent, which is \( \frac{1}{4} \). This slope is essential for forming the tangent line at that particular point on the curve.

Point-Slope Form

Once we have the slope from the derivative, the next step is to construct the tangent line's equation using the point-slope form. This form is particularly convenient when you know a point on the line and the slope.

The point-slope form of a line is given by:

• \( y - y_1 = m(x - x_1) \)

Here, \( m \) is the slope, and \((x_1, y_1)\) is the point through which the line passes. For the problem at hand:

• The slope \( m \) is \( \frac{1}{4} \).
• The point \((x_1, y_1)\) is \((1, 2)\).

Substituting these values into the point-slope form gives us:

• \( y - 2 = \frac{1}{4}(x - 1) \)

Simplifying the equation, you'll find:

• \( y = \frac{1}{4}x + \frac{7}{4} \)

This straightforward representation allows us to graphically see how the tangent line behaves relative to the curve, particularly at the point \((1, 2)\). The line touches the curve at just this one point, illustrating the definition of tangency.

Graphing Functions

Graphing can clarify abstract mathematical concepts by providing a visual component. It is the final step of the exercise, allowing us to see both the curve and the tangent line on a graph. Consider the function \( y = \sqrt{x+3} \). This function forms a curve that moves rightward from the vertical shift of \( y = \sqrt{x} \) due to the \(+3\).

To graph it:

• Recognize the domain: \( x \geq -3 \).
• The graph starts from the point where \( x = -3 \) and \( y = 0 \).
• It continues to rise gently as \( x \) increases.

Now, add the tangent line \( y = \frac{1}{4}x + \frac{7}{4} \):

• The line intersects the curve at \((1, 2)\).
• With the slope \( \frac{1}{4} \), it rises slowly as \( x \) moves from left to right.

Through graphing, both elements—the curve and the tangent—are exhibited, giving a better sense of the interplay between a function and its tangent line. This visual confirmation reinforces the solution as we see the curve and line meet precisely at \((1, 2)\). It solidifies understanding, linking derivative calculations to their geometric interpretation.