Question

Find the tangent line(s) to the curve \(y=x^{3}-9 x\) through the point (1,-9).

Short answer

The tangent line to the curve \(y = x^3 - 9x\) through the point (1, -9) has the equation \(y = -6x -3\).

Step-by-step solution

Verify the Point on the Curve

Substitute the coordinates of the point (1, -9) into the curve equation \(y = x^3 - 9x\). If the point lies on the curve, the left and right sides of the equation will be equal.

Calculate the Derivative to Find the Slope

The derivative of \(y = x^3 -9x\) is \(y' = 3x^2 -9\). Use this derivative formula to find the slope, m, of the tangent line at x=1.

Use Point-Slope Form to Find Equation of Line

Once we have the slope m, we can use the point-slope form of the line equation, \(y - y1 = m(x - x1)\), where (x1, y1) is the point on the curve and use the given point (1,-9) to get the equation of the tangent line.

Solve for y to Get the Equation of the Tangent Line

Solve the equation y - y1 = m(x - x1) for y to get the equation of the tangent line in slope-intercept form: \(y = mx + b\).

Key concepts

Derivative

When we talk about derivatives in calculus, we are discussing a fundamental concept that measures how a function changes as its input changes. This is akin to examining how quickly a car is moving at a specific moment if you think of the function as a path the car is driving along.

The formal definition of the derivative of a function at a point, represented as \( f'(x) \), is the limit of the average change (or difference quotient) as the interval considered approaches zero. For our example where the function is \( y = x^3 - 9x \), the derivative \( y' = 3x^2 - 9 \) represents the slope of the curve at any point \( x \). In simpler terms, if you were walking along the graph of the function, the derivative tells you exactly how steep the hill is at every step.

Slope of a Curve

Imagine you're taking a leisurely hike up a mountain, the path ahead of you twisting and turning. The slope of the curve in calculus resembles the incline or decline you experience on such a hike. It describes the direction and steepness of the curve at any given point.

Mathematically speaking, the slope at a specific point on a curve is the value of the derivative at that point. Therefore, when we calculated the derivative \( y' = 3x^2 - 9 \) for the function \( y = x^3 - 9x \) and found the slope \( m \) at \( x = 1 \), we discovered how steep the curve is at that precise location. On a graph, it shows how fast the \( y \) value is changing compared to the \( x \) value. The slope is a crucial piece of information because, in the context of finding tangent lines, it gives us the exact angle at which the tangent line should leave the curve.

Point-Slope Form

Sometimes, you need to pin a line to a specific point, like pinning a note to a bulletin board. The point-slope form does just that for lines on a graph. It is an equation of a line that includes both the line's slope and its intersection at a particular point. Expressed as \( y - y1 = m(x - x1) \) where \( m \) stands for the slope and \( (x1, y1) \) is a given point through which the line passes.

For our initial problem, once we have the slope \( m \) from the function's derivative at \( x = 1 \) and our point (1, -9), we plug these into the point-slope equation. Algebra then takes the wheel, allowing us to isolate \( y \) to find the line's equation in slope-intercept form (\( y = mx + b \) ). This makes plotting the line straightforward as it reveals both the incline and the starting point of the line on the \( y \) axis, serving as a 'map' to draw the line directly on the graph.