write an equation for the line tangent to the curve at the point (1, 3)


find the value of (d^2y)/(dx^2) at the point (1,3) show all work

To find the equation of the tangent line to a curve at a given point and the second derivative \( \frac{d^2y}{dx^2} \), we'll follow these steps:

### 1. **Step 1: Define the curve equation**

We assume you have a curve in mind, but it is not specified. Let's assume a general curve like:
\[
y = f(x)
\]
You can specify the curve if needed, but for now, I'll proceed with an arbitrary one to demonstrate the method.

### 2. **Step 2: Find the equation of the tangent line at \( (1, 3) \)**

To find the tangent line at a point \( (x_1, y_1) = (1, 3) \), we need to:
- Compute the first derivative \( \frac{dy}{dx} \) of the function \( y = f(x) \).
- Evaluate the slope of the tangent line at \( x = 1 \) (i.e., find \( \frac{dy}{dx} \bigg|_{x=1} \)).

The equation of the tangent line is given by the point-slope form of a line:
\[
y - y_1 = m(x - x_1)
\]
where \( m \) is the slope of the tangent line, which is \( \frac{dy}{dx} \) at \( x_1 = 1 \).

### 3. **Step 3: Find the second derivative \( \frac{d^2y}{dx^2} \) at \( (1, 3) \)**

To find the concavity or acceleration of the curve at \( x = 1 \), we compute the second derivative \( \frac{d^2y}{dx^2} \) and evaluate it at \( x = 1 \).

### Example:

Let's use a specific example to illustrate the process. Suppose the curve is:

\[
y = x^3 + 2x^2 + x
\]

#### 3.1. **First derivative \( \frac{dy}{dx} \):**
Differentiate the function with respect to \( x \):
\[
\frac{dy}{dx} = 3x^2 + 4x + 1
\]
#### 3.2. **Find the slope at \( x = 1 \):**
Evaluate the derivative at \( x = 1 \):
\[
\frac{dy}{dx} \bigg|_{x=1} = 3(1)^2 + 4(1) + 1 = 3 + 4 + 1 = 8
\]
So, the slope of the tangent line at \( x = 1 \) is \( 8 \).

#### 3.3. **Equation of the tangent line:**
Using the point-slope form \( y - 3 = 8(x - 1) \), we get:
\[
y - 3 = 8(x - 1)
\]
Simplify the equation:
\[
y = 8x - 8 + 3 \quad \Rightarrow \quad y = 8x - 5
\]
Thus, the equation of the tangent line is:
\[
y = 8x - 5
\]

### 4. **Find the second derivative \( \frac{d^2y}{dx^2} \):**
Differentiate \( \frac{dy}{dx} = 3x^2 + 4x + 1 \) again with respect to \( x \):
\[
\frac{d^2y}{dx^2} = 6x + 4
\]
Evaluate the second derivative at \( x = 1 \):
\[
\frac{d^2y}{dx^2} \bigg|_{x=1} = 6(1) + 4 = 6 + 4 = 10
\]

### Final Results:
- **Equation of the tangent line** at \( (1, 3) \): \( y = 8x - 5 \).
- **Second derivative** at \( x = 1 \): \( \frac{d^2y}{dx^2} = 10 \).

---

Would you like more details on any step or have any questions?

### Relative questions:
1. How do you find the derivative of a product of functions?
2. What is the geometric significance of the second derivative?
3. Can the tangent line be horizontal, and under what conditions?
4. How would you find the normal line to the curve at a point?
5. How does the tangent line relate to the concavity of a curve?

### Tip:
The second derivative tells you about the concavity of the function. If \( \frac{d^2y}{dx^2} > 0 \), the function is concave up; if \( \frac{d^2y}{dx^2} < 0 \), it's concave down.

write an equation for the line tangent to the curve at the point (1, 3) and find the value of (d^2y)/(dx^2) at the point (1,3) show all work

Discover how to find the equation of the tangent line to a curve at the point (1, 3) and compute the second derivative at the same point. This step-by-step solution involves differentiation, including the first and second derivatives. Suitable for students in Grades 11-12.

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