
*y* = *x*2 + 6*x*

 at the point 

(−2, −8)

 by using the following parameters.

It looks like you're asking to evaluate or analyze the function \( y = x^2 + 6x \) at the point \( (-2, -8) \) using some parameters, though the parameters weren't mentioned in your message.

We can proceed with the following steps if the task is to confirm whether \( (-2, -8) \) is on the curve, or find derivatives, tangent lines, or other parameters at that point. Here's a breakdown:

### Step 1: Check if \( (-2, -8) \) is on the curve.
We'll plug \( x = -2 \) into the equation and see if the result equals \( y = -8 \).

Given:
\[
y = x^2 + 6x
\]
Substitute \( x = -2 \):

\[
y = (-2)^2 + 6(-2)
= 4 - 12
= -8
\]

So, the point \( (-2, -8) \) is on the curve.

### Step 2: Derivative of the function \( y = x^2 + 6x \)
The derivative represents the slope of the tangent line at any point on the curve.

Using basic rules of differentiation:
\[
\frac{dy}{dx} = 2x + 6
\]
Now substitute \( x = -2 \) into the derivative:

\[
\frac{dy}{dx} = 2(-2) + 6 = -4 + 6 = 2
\]

Thus, the slope of the tangent line at \( (-2, -8) \) is 2.

### Step 3: Equation of the tangent line at \( (-2, -8) \)
The slope of the tangent line is 2, and it passes through the point \( (-2, -8) \). The equation of the tangent line is given by:

\[
y - y_1 = m(x - x_1)
\]
Where \( m \) is the slope and \( (x_1, y_1) = (-2, -8) \).

Substitute the values:

\[
y - (-8) = 2(x - (-2))
\]

Simplifying:

\[
y + 8 = 2(x + 2)
\]
\[
y + 8 = 2x + 4
\]
\[
y = 2x - 4
\]

Thus, the equation of the tangent line at \( (-2, -8) \) is \( y = 2x - 4 \).

### Summary:
1. The point \( (-2, -8) \) is on the curve.
2. The derivative (slope of the tangent line) at \( (-2, -8) \) is 2.
3. The equation of the tangent line at \( (-2, -8) \) is \( y = 2x - 4 \).

Would you like more details on any of the steps or have other questions?

Here are five related questions:
1. How do we find the second derivative of the function?
2. How do we calculate the normal line at \( (-2, -8) \)?
3. What is the concavity of the function near the point \( (-2, -8) \)?
4. How do we find the y-intercept of the tangent line?
5. What is the slope of the curve at different values of \( x \)?

**Tip**: The first derivative gives the slope of the tangent line, while the second derivative reveals the concavity (how the curve bends) at a point.

y = x^2 + 6x at the point (−2, −8) by using the following parameters.

Explore how to evaluate the function y = x^2 + 6x at the point (-2, -8) and find the tangent line. This solution covers differentiation, checking points on a curve, and deriving the equation of the tangent line. Ideal for students in Grades 10-12.

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