Question

Horizontal Tangent At what point is the tangent to \(f(x)=x^{2}+4 x-1\) horizontal?

Short answer

The tangent to the function \(f(x)=x^{2}+4 x-1\) is horizontal at the point (-2, -5).

Step-by-step solution

Calculate the derivative

The derivative of the function \(f(x) = x^{2} + 4x -1\) is calculated using the power rule which states that the derivative of \(x^{n}\) is \(n x^{n-1}\). Therefore, the derivative \(f'(x) = 2x + 4\).

Solve the derivative equal to 0

To find the x-coordinate where the function has a horizontal tangent, the derivative should be set to zero. So, solve for x in the equation \(f'(x) = 0\). This results in x being equal to -2.

Find the y-coordinate

Now, to get the corresponding y-coordinate at this point of tangency, substitute x equal to -2 into the original function \(f(x) = x^{2} + 4x -1\). So, \(f(-2) = (-2)^{2} + 4(-2) - 1 = 4 - 8 - 1 = -5\).

Key concepts

Derivative Calculation

Understanding the process of derivative calculation is essential when it comes to analyzing the behavior of functions. A derivative represents the rate at which a function's output changes as its input changes. Essentially, it gives us the slope of the function at any given point.

To calculate a derivative, you need to understand the limits and the concept of differentiability. For most algebraic functions, we can apply various differentiation rules to simplify the process. In the given exercise, finding a horizontal tangent requires us to first determine the derivative of the function.

Power Rule Differentiation

One of the most widely used differentiation techniques is the power rule. This rule applies to functions where the variable is raised to a power. The power rule states that to find the derivative of a function in the form of x^n, you multiply the exponent by the coefficient and then decrease the exponent by one.

In mathematical notation, if we have a function f(x) = ax^n, its derivative f'(x) is anx^(n-1). Applying this to the provided exercise, for the function f(x) = x^2 + 4x - 1, we take the derivative of each term separately - the x^2 becomes 2x and the 4x just 4 (as the derivative of a linear term ax is just a). This is how we arrive at the derivative f'(x) = 2x + 4.

Tangent Line Problem

A tangent line problem involves finding the equation of the line that just 'touches' a curve at a given point without crossing it. The slope of this tangent line at any given point is exactly the value of the derivative of the function at that point.

However, when the tangent is horizontal, the slope is 0. Consequently, to find where this occurs, we set the derivative equal to zero and solve for x. In the context of our exercise, solving f'(x) = 0 gives us the x-coordinate of the point where the tangent is horizontal. Lastly, we find the corresponding y-coordinate by substituting our x-value back into the original function, which completes the solution to the tangent line problem.