Problem 22 - Calculus for Biology and Medicine

Question

Use the formal definition to find the derivative of \(y=\) \(-2 x^{2}\) at \(x=1\) (b) Show that the point \((1,-2)\) is on the graph of \(y=-2 x^{2}\), and find the equation of the tangent line at the point \((1,-2)\). (c) Graph \(y=-2 x^{2}\) and the tangent line at the point \((1,-2)\) in the same coordinate system.

Step-by-Step Solution

Answer

The derivative at \(x = 1\) is -4; the tangent line is \(y = -4x + 2\).

Step 1: Understanding the Function

The function provided is \( y = -2x^2 \). This is a quadratic function, and we are interested in finding its derivative at a specific point \( x = 1 \).

Step 2: Finding the Derivative using Definition

The derivative of a function \( f(x) \) at a point \( x \) is defined as \( \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \). Here, \( f(x) = -2x^2 \). We find \( f(x+h) = -2(x+h)^2 = -2(x^2 + 2xh + h^2) = -2x^2 - 4xh - 2h^2 \). Substituting in the definition: \( \lim_{h \to 0} \frac{(-2x^2 - 4xh - 2h^2) + 2x^2}{h} = \lim_{h \to 0} \frac{-4xh - 2h^2}{h} \). Simplifying gives us \( \lim_{h \to 0} (-4x - 2h) = -4x \). So, the derivative is \(-4x\).

Step 3: Evaluate Derivative at Specific Point

Substitute \( x = 1 \) into the derivative \(-4x\). Thus, \( f'(1) = -4 \times 1 = -4 \). So, the derivative at \( x = 1 \) is -4.

Step 4: Confirm Point on Graph

To confirm that the point \((1, -2)\) is on the graph of \( y = -2x^2 \), substitute \( x = 1 \) into the original equation: \( y = -2 \times 1^2 = -2 \). Hence, the point \( (1, -2) \) is indeed on the graph.

Step 5: Find Tangent Line Equation

The equation of a tangent line is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) is the point of tangency. Here, \( m = -4 \), \( x_1 = 1 \), and \( y_1 = -2 \). Substituting, \( y + 2 = -4(x - 1) \), which simplifies to \( y = -4x + 4 - 2 = -4x + 2 \). So, the equation is \( y = -4x + 2 \).

Step 6: Graph the Function and Tangent Line

Graph the function \( y = -2x^2 \) which is a downward opening parabola. Next, graph the tangent line equation \( y = -4x + 2 \) which intersects the parabola at the point \( (1, -2) \). This depiction visually confirms the point of tangency and the slope.

Key Concepts

Tangent Line

A tangent line at a point on a curve is a straight line that just "touches" the curve at that point. Think of it as a line that brushes against the curve without crossing it at that specific location. For the equation of a tangent line, you need to know:

• The slope of the line at the given point, which is the derivative of the function at that point.
• The point of tangency, (x_1, y_1), on the curve.

The formula to find the tangent line is:\[ y - y_1 = m(x - x_1) \]where \(m\) is the slope found from the derivative.

In our example, the function is \( y = -2x^2 \) and its derivative is \(-4x\). At the point \(x = 1\), the slope \(m\) is \(-4\). Thus, the tangent line at the point (1, -2) can be found by substituting into the formula: \[ y + 2 = -4(x - 1) \]which simplifies to \[ y = -4x + 2 \].This line grazes the parabola at exactly the point \((1, -2)\).

Quadratic Function

Quadratic functions are a type of polynomial with a degree of 2. Their general form is:\[ y = ax^2 + bx + c \]In this form, \(a\), \(b\), and \(c\) are constants, and the highest power of \(x\) is 2. The graph of a quadratic function is a parabola.

• If \(a > 0\), the parabola opens upward.
• If \(a < 0\), as in our example \(y = -2x^2\), the parabola opens downward.

Parabolas are symmetric around their vertex, which is their peak or lowest point. For \(y = -2x^2\), the vertex is at the origin (0, 0), and it opens downward, creating a U-shaped curve.

Quadric functions are useful in understanding the vertex, axis of symmetry, and directional opening of the graph. In our example, the negative \(a\) value indicates the parabola flips and opens downward.

Formal Definition of Derivative

The formal definition of a derivative is used to calculate the rate at which a function is changing at any given point. This notion forms the foundation of differential calculus.

In simple terms, if you have a function \(f(x)\), the derivative of the function at any point \(x\) is given by:\[\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]This limit calculates the slope of the tangent line to the curve at point \(x\).

• The closer \(h\) gets to 0, the more accurate the slope of the tangent line becomes.
• This expression gives us the instantaneous rate of change of the function.

For the function \( y = -2x^2 \), we calculated the derivative using this formal definition and found it to be \(-4x\). Evaluating at \(x = 1\), we found that the slope of the tangent line is -4. This illustrates how derivatives help in understanding not just the value but the behavior of functions at specific points.