To find the equations of the tangent lines for each function at the given points, we'll follow these steps:

1. Find the derivative of the function to determine the slope of the tangent line.
2. Evaluate the derivative at the given point to find the slope at that specific point.
3. Use the point-slope form of a line equation and rewrite it in slope-intercept form.

Let's go through each problem step-by-step.

Problem 1

Function: $y = x^3 - 3x^2 + 2$ Point: $(3, 2)$

Step 1: Find the derivative

The derivative of $y = x^3 - 3x^2 + 2$ is: $y' = 3x^2 - 6x$

Step 2: Evaluate the derivative at $x = 3$

Substitute $x = 3$ into $y'$ to find the slope at the point $(3, 2)$: $y'(3) = 3(3)^2 - 6(3) = 3(9) - 18 = 27 - 18 = 9$ So, the slope of the tangent line at $(3, 2)$ is 9.

Step 3: Write the equation of the tangent line

Using the point-slope form of a line: $y - y_1 = m(x - x_1)$ $y - 2 = 9(x - 3)$ Expanding and simplifying to slope-intercept form: $y = 9x - 27 + 2$ $y = 9x - 25$ Thus, the equation of the tangent line is: $y = 9x - 25$

Problem 2

Function: $y = -\frac{5}{x^2 + 1}$ Point: $\left(-1, -\frac{5}{2}\right)$

Step 1: Find the derivative

Rewrite the function as $y = -5(x^2 + 1)^{-1}$ to apply the power rule. $y' = 5 \cdot (x^2 + 1)^{-2} \cdot (2x) = \frac{10x}{(x^2 + 1)^2}$

Step 2: Evaluate the derivative at $x = -1$

Substitute $x = -1$ into $y'$ to find the slope at the point $\left(-1, -\frac{5}{2}\right)$: $y'(-1) = \frac{10(-1)}{((-1)^2 + 1)^2} = \frac{-10}{(1 + 1)^2} = \frac{-10}{4} = -\frac{5}{2}$ So, the slope of the tangent line at $\left(-1, -\frac{5}{2}\right)$ is $-\frac{5}{2}$.

Step 3: Write the equation of the tangent line

Using the point-slope form of a line: $y - y_1 = m(x - x_1)$ $y - \left(-\frac{5}{2}\right) = -\frac{5}{2}(x + 1)$ Simplifying: $y + \frac{5}{2} = -\frac{5}{2}x - \frac{5}{2}$ Subtract $\frac{5}{2}$ from both sides: $y = -\frac{5}{2}x - 5$ Thus, the equation of the tangent line is: $y = -\frac{5}{2}x - 5$

Summary of Tangent Line Equations

1. For $y = x^3 - 3x^2 + 2$ at $(3, 2)$: $y = 9x - 25$
2. For $y = -\frac{5}{x^2 + 1}$ at $\left(-1, -\frac{5}{2}\right)$: $y = -\frac{5}{2}x - 5$

Would you like further explanation on any of these steps, or do you have any other questions?

Follow-up Questions:

1. How can you find the slope of a tangent line for any given point on a curve?
2. What does the derivative of a function tell us about the behavior of the function at a specific point?
3. Why is it important to rewrite functions in a suitable form before taking derivatives?
4. How can you verify that the tangent line equation is correct?
5. How does the slope-intercept form of a line relate to the general concept of linear equations?

Tip:

When finding the tangent line, always confirm the function's form is differentiable at the point in question to ensure a valid tangent exists.