Problem 6 - Precalculus: Mathematics for Calculus

Question

Find the slope of the tangent line to the graph of \(f\) at the given point. $$f(x)=1+2 x-3 x^{2}, \quad \text { at }(1,0)$$

Step-by-Step Solution

Answer

The slope is -4.

Step 1: Find the Derivative

The slope of the tangent line to a function at a given point is the value of the derivative at that point. Compute the derivative of \(f(x) = 1 + 2x - 3x^2\). The derivative of a function \(f(x)\) is found by differentiating each term separately. The derivative of \(1\) is \(0\), of \(2x\) is \(2\), and for \(-3x^2\) it is \(-6x\). Therefore, the derivative \(f'(x)\) is \(2 - 6x\).

Step 2: Substitute the Given Point into the Derivative

Now, we need to find the slope of the tangent line at the point \((1, 0)\). Substitute \(x = 1\) into the derivative \(f'(x) = 2 - 6x\). So, \(f'(1) = 2 - 6 \times 1 = 2 - 6 = -4\).

Step 3: Interpret the Result

The value obtained is the slope of the tangent line to the graph of \(f\) at the given point \((1,0)\). Therefore, the slope is \(-4\).

Key Concepts

Slope of Tangent Line

To understand the slope of a tangent line, imagine a curve. At a point on this curve, the tangent line "just touches" without crossing it. The slope of this line is crucial because it tells us how steep the curve is at that point. To find this slope, we can use the derivative of the function. The derivative essentially gives us a formula to calculate slope at any point on the curve.
By plugging the x-value of the given point into this derivative, we get the slope at that exact location. For example, with the function given in the exercise, plugging the x-value into the derivative gives us a slope of -4 at the point \((1, 0)\). This means the tangent line is descending steeply at that point on the curve.

Function Differentiation

Differentiation is the process of finding the derivative of a function. The derivative shows how a function changes as its input changes. Think of it like understanding the pace at which a position changes over time. For the quadratic function \(f(x) = 1 + 2x - 3x^2\), differentiating each term separately simplifies the process.

• The constant \(1\) has a derivative of \(0\), because it doesn’t change.
• The term \(2x\) changes at a constant rate, so its derivative is \(2\).
• The term \(-3x^2\) has a derivative of \(-6x\), using the power rule.

Combining these gives us the derivative \(f'(x) = 2 - 6x\). This derivative can now be used to find the slope of the tangent line, reflecting how the function behaves near any given point.

Finding Derivatives

Finding derivatives is a key part of working with functions in calculus. It involves calculating how quickly a function's output is changing at any given point, which is foundational for understanding the behavior of graphs.
To find the derivative, we typically apply rules such as the power rule, product rule, or chain rule, depending on the function's complexity. For basic polynomials, like in our example, the power rule suffices. This rule states that for any term \(ax^n\), the derivative is \(nax^{n-1}\).
Using the function \(f(x) = 1 + 2x - 3x^2\), we found:

• The derivative of the constant \(1\) is \(0\).
• For \(2x\), the derivative is \(2\).
• For the quadratic term \(-3x^2\), applying the power rule, turns into \(-6x\).

So, the derivative \(f'(x) = 2 - 6x\) provides a complete picture of the slope at any point on the graph.