Question

Finding an Equation of a Tangent Line In Exercises \(33-38,\) find an equation of the line that is tangent to the graph of \(f\) and parallel to the given line. $$\begin{array}{ll}{\text { Function }} & {\text { Line }} \\ {f(x)=2 x^{2}} & {4 x+y+3=0}\end{array}$$

Short answer

The equation of the line that is tangent to the graph of the function and parallel to the given line is \(y = -4x - 2\).

Step-by-step solution

Find the Slope of the Given Line

Rearrange the line equation \(4x + y + 3 = 0\) to a more familiar form, i.e. \(y = mx + b\), to find its slope. This yields \(y = -4x - 3\), so the slope of our given line is -4.

Differentiate the Function

The derivative of the function \(f(x) = 2x^2\) is \(f'(x) = 4x\). This represents the slope of the tangent line at any point x.

Find the Value of x Where the Slope of the Tangent Line is Same

To find the value of x for which the slope of the function equals the slope of the line, set \(f'(x)\) equal to the slope of the line and solve for x. So we get \(4x = -4\), this yields \(x = -1\).

Find the Corresponding y-coordinate

Plug \(x = -1\) back into the original function \(f(x) = 2x^2\), we get \(f(-1) = 2(-1)^2 = 2\), hence the point of tangency is \((-1, 2)\)

Find the Equation of the Tangent Line

Use the slope of the line and the point \((-1, 2)\) in the point-slope form of a line \(y - y1 = m(x - x1)\) to find the equation of the tangent line. This yields \(y - 2 = -4(x + 1)\), rearranging this gives us the final equation \(y = -4x - 2\).

Key concepts

Slope of a Line

The slope of a line is a measure of its steepness and is a key concept in understanding linear functions and their graphs. It represents the rate at which the y-coordinate of a point changes with respect to its x-coordinate along the line. To find the slope of a line represented by an equation like 4x + y + 3 = 0, we rearrange it into slope-intercept form, which is y = mx + b, where m is the slope. For the given equation, when rearranged to y = -4x - 3, the slope is -4. This indicates that for every unit increase in x, y decreases by four units.

Slope is commonly denoted by m and can be calculated between any two points (x1, y1) and (x2, y2) on the line as m = (y2 - y1) / (x2 - x1). Slope is a foundational element when discussing calculus and linear algebra, as it is closely related to rates of change and linear equations.

Derivative of a Function

A derivative of a function is a concept from calculus that measures how a function's output changes as its input changes. It is equivalent to the slope of the tangent line at any given point on the graph of the function. For instance, the derivative of f(x) = 2x^2 is found by applying rules of differentiation, resulting in f'(x) = 4x. This derivative function now gives us the slope of the tangent at any point x on the curve described by f(x).

In the context of our problem, we use the derivative to find the slope of the tangent line that is parallel to a given line. Understanding derivatives is essential for fields such as physics, engineering, and economics, where change is a constant factor.

Point-Slope Form

The point-slope form is one of the standard forms for writing the equation of a line. It's particularly useful when we have a point on the line and the slope. The general formula is y - y1 = m(x - x1), where (x1, y1) is the point on the line, and m is the slope. This form allows us to write the equation of a line quickly given this information.

For example, if the slope of the line is -4 and the point of tangency is (-1, 2), we can directly substitute these values into the formula to get y - 2 = -4(x + 1). The point-slope form is particularly handy when dealing with the equation of a tangent line in calculus, as it smoothly incorporates the coordinates of the tangency point and the slope provided by the derivative.

Tangency Point

A tangency point is where a line touches a curve exactly at one point and does not cross the curve. At this exact point, the slope of the curve, given by the derivative, is equal to the slope of the line. Thus, finding the tangency point is a key step in determining the equation of a tangent line.

In the given exercise, by setting the derivative f'(x) = 4x equal to the slope of the given line, -4, we are essentially finding the x-coordinate where the curve has the same slope as the line. Solving 4x = -4 gives us the x-coordinate of the tangency point, which is -1. We then find the y-coordinate by substituting x into the original function f(x). Thus, the tangency point in our problem is (-1, 2), which is crucial for writing the equation of the tangent line.