Question

Find an equation of the line that is tangent to the graph of \(f\) and parallel to the given line. $$ f(x)=x^{2}+1 \quad 2 x+y=0 $$

Short answer

The equation of the line tangent to the function \(f(x) = x^{2} + 1\) and parallel to the given line \(2x + y = 0\) is \(y = -2x\).

Step-by-step solution

Determine the derivative of the function

The derivative of the function \(f(x)\) helps understand the behaviour of the function. The derivative of \(f(x) = x^{2} + 1\) is \(f'(x) = 2x\).

Determine the x-coordinate for the point of tangency

The derivative \(f'(x) = 2x\) gives the slope of the tangent line for a given x-value. We need this slope to be -2 (as the slopes of parallel lines are equal). Hence, solving the equation \(2x = -2\), we get \(x = -1\)

Determine the y-coordinate for the point of tangency

The y-coordinate of the point where the line is tangent can be obtained by substituting the x-value derived previously into the function \(f(x)\). Hence \(f(-1) = (-1)^{2} + 1 = 2\)

Formulate the equation of the line with the derived x and y coordinates

The equation of the line is written in the slope-intercept form: \(y = mx + c\), where \(m\) is the slope and \(c\) is the y-intercept. We know \(m = -2\) and the point of tangency is (-1, 2). Plugging these into the equation gives the equation of the line: \(y = (-2)(x + 1) + 2 = -2x\)

Key concepts

Derivative of a Function

The derivative of a function is like a snapshot of its behaviour at any given point—it tells us how the function is changing in that instant. Think of it as a mathematical way to measure the 'speed' at which the function values are increasing or decreasing as we move along the graph.

For example, the derivative of the function in our exercise, which is \(f(x) = x^{2} + 1\), is \(f'(x) = 2x\). This formula represents the slope of the tangent line to the graph of the function at any point \(x\). If you plug any value of \(x\) into \(f'(x)\), you get the slope of the tangent line at that particular \(x\) on the graph. It's like knowing the steepness of a hill at different points just by looking at its formula.

Slope of a Tangent Line

The slope of a tangent line to a function at a given point is the value of the derivative at that point. Imagine drawing a straight line that just skims a curve on the graph. This line is the tangent, and its slope tells us how steep the line is at the point where it touches the curve.

In our exercise, we used the derivative \(f'(x) = 2x\) to find the slope of the tangent line that should be parallel to the given line with the slope of -2. By setting \(2x = -2\), we are saying 'find us the exact spot on the curve where the tangent line is just as steep as the given line'. Therefore, for the derivative equal to -2, the matching point on the x-axis is \(x = -1\), which is exactly what we needed for our solution.

Slope-Intercept Form of a Line

When it comes to writing down the equation of a line, the slope-intercept form is like a quick recipe with just two ingredients: the slope \(m\) and the y-intercept \(c\). The form looks like \(y = mx + c\), where \(y\) and \(x\) are the coordinates of any point on the line.

Using the values we've found, our function's tangent line has the slope \(m = -2\) and goes through the point \((-1, 2)\). To find the y-intercept \(c\), we just put the slope and the point into the slope-intercept formula. So our line equation becomes \(y = -2x\), as we found after plugging in the slope and the coordinates of the point where the line and the function meet. This form is extremely handy because it allows anyone to quickly sketch the line or determine if a point lies on it, just by using the slope and the y-intercept.