Problem 58 - Precalculus with Limits

Question

In Exercises 55-58, find an equation of the line that is tangent to the graph of and parallel to the given line. \(\quad \quad Function \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad Line\) \(f(x) = x^2 - x \quad \quad \quad \quad \quad \quad \quad \quad x+2y-6=0\)

Step-by-Step Solution

Answer

Hence, the equation of the line that is tangent to the graph of \(f(x) = x^2 - x\) and parallel to the line \(x+2y-6=0\) is \(y = -0.5x + 0.1875.\)

Step 1: Finding the derivative

Firstly, compute the derivative \(f'(x)\) of the function \(f(x) = x^2 - x\). This yields \(f'(x) = 2x - 1\). The derivative at any point gives the slope of the tangent line at that point.

Step 2: Finding the slope of the given line

Rearranging the given straight line equation \(x+2y–6=0\) to slope-intercept form to find the slope. This gives us \(y= -0.5x +3\), so the slope of the line is -0.5.

Step 3: Finding the x-coordinate of the tangent point

To find the x-coordinate of the point where the tangent line touches the function, set \(f'(x)\) equal to the slope of the given straight line, -0.5. This gives the equation \(2x - 1 = -0.5\), which we solve to yield \(x = 0.25\).

Step 4: Finding the y-coordinate of the tangent point

Substitute \(x=0.25\) into \(f(x)\) to find the corresponding y-coordinate of the point of tangency. This gives \(y=0.25^2-0.25= -0.0625.\)

Step 5: Finding the equation of the tangent line

Lastly, using the point-slope form of a linear equation, \(y - y_1 = m(x - x_1)\), where \(m=-0.5\) is the slope and \((x_1, y_1)= (0.25, -0.0625)\) is the point of tangency, the equation of the tangent line is found to be \(y+0.0625 = -0.5(x-0.25)\). Rearranging this gives \(y = -0.5x + 0.1875.\)

Key Concepts

Derivatives

In the realm of calculus, derivatives stand as one of the fundamental concepts. They are used to describe the rate at which a function's value changes at a certain point. Essentially, when you find a derivative of a function such as f(x) = x^2 - x, you're determining the slope of the tangent line to the graph of the function at any point 'x'.

Imagine driving on a hilly road; the derivative tells you how steep the hill is at any instant. It’s like the function is your path, and the derivative is your speedometer showing how fast the elevation is changing. In the context of our exercise, the derivative is calculated as f'(x) = 2x - 1, which means for any value of 'x', the slope of the tangent at that point is given by this derivative function.

Moreover, when you're trying to find where a function is flat (where the slope is zero), or where it's increasing or decreasing the fastest, derivatives become an indispensable tool. They bring the graphical behavior of functions into the algebraic realm, offering a bridge between the visual and numerical worlds of mathematics.

Slope-Intercept Form

When you're looking at a line on a graph, the slope-intercept form of its equation is incredibly useful to quickly understand its characteristics. This form is written as y = mx + b where 'm' represents the slope of the line and 'b' is the y-intercept, the point where the line crosses the y-axis.

To visualize this, suppose you're plotting your monthly savings on a graph. The 'slope' shows how quickly you're saving over time, and the 'intercept' shows how much money you had saved initially before you started saving monthly. In this manner, the slope-intercept form gives you a direct look at how steeply the line rises or falls and where it begins.

In our exercise, we rearranged the line's equation to the slope-intercept form, which yielded y = -0.5x + 3. The coefficient of 'x' (-0.5) here is the slope of the line and '3' is where the line crosses the y-axis. This quick algebraic manipulation shows us how the graph behaves without needing to plot it.

Point-Slope Form

Another incredibly useful form of a linear equation is the point-slope form, particularly when you know a point on the line and the slope. This is given by y - y_1 = m(x - x_1), where (x_1, y_1) is the known point and 'm' is the slope of the line.

This is akin to having a specific destination in mind and knowing the direction you need to go. The point (x_1, y_1) represents your starting location and the slope 'm' tells you which direction and how steeply to head towards your destination. It's a straightforward way to write the equation of a line when you have specific information about its slope and one point it passes through.

In the exercise we're tackling, using the calculated slope of the tangent line and the point of tangency, the point-slope form gives us the precise equation of the tangent line. The equation y+0.0625 = -0.5(x-0.25) is derived from knowing these particular details about our tangent line. This form is the go-to method when the line's behavior is revealed through a point and a direction rather than through its overall intercept with axes.