Question

Finding an Equation of a Tangent Line In Exercises 81 and \(82,\) find an equation of the tangent line to the graph of the function \(f\) through the point \(\left(x_{0}, y_{0}\right)\) not on the graph. To find the point of tangency \((x, y)\) on the graph of \(f,\) solve the equation $$ f^{\prime}(x)=\frac{y_{0}-y}{x_{0}-x} $$ $$ \begin{array}{l}{f(x)=\sqrt{x}} \\ {\left(x_{0}, y_{0}\right)=(-4,0)}\end{array} $$

Short answer

The equation of the tangent line to the function \(f(x) = \sqrt{x}\) through the point (-4,0) is \(y = \frac{x}{5} + \frac{16}{5}\).

Step-by-step solution

Finding the derivative of the function

Firstly, we need to find the derivative of the function \(f(x) = \sqrt{x}\), that is \(f^{\prime}(x)\). We know from the power rule, the derivative of \(x^n\) is \(nx^{n-1}\). Hence, if we rewrite \(\sqrt{x}\) as \(x^{1/2}\), we can apply the power rule: \(f^{\prime}(x) = \frac{1}{2}x^{-1/2}\). This simplifies to \(f^{\prime}(x) = \frac{1}{2\sqrt{x}}\).

Applying the given equation

Next, we need to substitute the value of \(f^{\prime}(x)\) into the given equation \(f^{\prime}(x) = \frac{y_0 - y}{x_0 - x}\). Upon substitution, we get \(\frac{1}{2\sqrt{x}} = \frac{0 - y}{-4 - x}\). We can simplify this further to \(2\sqrt{x} = \frac{y}{x+4}\). Cross multiplication gives \(y = 2\sqrt{x}(x+4)\).

Finding the coordinates of the point

The expression derived in step 2, \(y = 2\sqrt{x}(x+4)\), represents a family of tangent lines. The actual tangent line can be found by substituing \(y = \sqrt{x}\), since point \((x, y)\) lies on the graph of \(f(x)\). So, we substitute to get \(2\sqrt{x}(x+4) = \sqrt{x}\). Solving this gives \(x = 16\). Thus, \(y = \sqrt{16} = 4\). So, the point of tangency is \((16, 4)\).

Equation of the tangent line

We know the equation of a line passing through \((x_1, y_1)\) and \((x_2, y_2)\) can be given as \((y - y_1) = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1)\). Substituting the point of tangency \((16, 4)\) as \((x_1, y_1)\) and the point \((-4, 0)\) as \((x_2, y_2)\), we can write the equation of the tangent line as \((y - 4) = \frac{4 - 0}{16 - (-4)} (x - 16)\). Simplifying this gives the equation of the line as \(y = \frac{x}{5} + \frac{16}{5}\).

Key concepts

Derivative

To find the equation of a tangent line, we start by understanding the derivative. The derivative, often denoted by \(f'(x)\), represents the slope of the tangent line to the graph of a function at a specific point. It tells us how the function changes as \(x\) changes.
The function given is \(f(x) = \sqrt{x}\), which can be rewritten using exponents as \(x^{1/2}\). Using the power rule, which allows us to differentiate power functions, we find the derivative. The power rule states that if \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\).
Applying the power rule, we differentiate \(x^{1/2}\) to get \(f'(x) = \frac{1}{2}x^{-1/2}\). This simplifies further to \(\frac{1}{2\sqrt{x}}\).

• This derivative tells us the slope of the tangent line at any point \(x\).

Point of Tangency

The point of tangency is where the tangent line touches the curve. This is a crucial part of finding the tangent line's equation.
Given the problem, the point \((x_0, y_0) = (-4, 0)\) is not on the graph of \(f\). Thus, we need to determine the point \((x, y)\) on the graph where the line will be tangent.
We use the equation \(f'(x) = \frac{y_0 - y}{x_0 - x}\) to find the point of tangency. After substituting \(f'(x) = \frac{1}{2\sqrt{x}}\) into this equation, and simplifying, we find that \(x = 16\) and subsequently \(y = 4\).

• At this point \((16, 4)\), the line will just touch the curve.

Equation of a Line

Now that we have the point of tangency, we can find the equation of the tangent line. The formula for the equation of a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) is:
\((y - y_1) = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1)\).
Substitute \((16, 4)\) as \((x_1, y_1)\) and \((-4, 0)\) as \((x_2, y_2)\). This results in:
\((y - 4) = \frac{4 - 0}{16 - (-4)} (x - 16)\).
Solving this gives us the equation \(y = \frac{x}{5} + \frac{16}{5}\).

• This equation represents the tangent line to the curve \(f(x)\) at the point \((16, 4)\).

Power Rule

The power rule is a fundamental tool in calculus for finding derivatives. It is especially useful when dealing with polynomial or power functions, like \(f(x) = x^n\).
The rule states that the derivative of \(x^n\) is \(nx^{n-1}\).
In this task, we applied the power rule to the function \(f(x) = \sqrt{x}\). By rewriting \(\sqrt{x}\) as \(x^{1/2}\), we made it suitable for differentiation using the power rule.

• This resulted in the derivative \(f'(x) = \frac{1}{2}x^{-1/2}\), which is crucial for finding the slope at any given point.