https://file.aiquickdraw.com/mathbot/file/1727033595986gpos3vue.jpg

Let's break down and solve the problem from the image.

### Part 1: Finding \( h'(3) \)
You are given the function \( h(x) = 2 - 4x^3 \). To find \( h'(3) \), we first need to calculate the derivative \( h'(x) \), and then substitute \( x = 3 \).

The derivative of \( h(x) = 2 - 4x^3 \) is:
\[
h'(x) = \frac{d}{dx} \left(2 - 4x^3\right) = -12x^2
\]

Now substitute \( x = 3 \) into this derivative:
\[
h'(3) = -12(3)^2 = -12(9) = -108
\]

Thus, \( h'(3) = -108 \).

### Part 2: Equation of the Tangent Line
The equation of the tangent line is given by the formula:
\[
y = mx + b
\]
where \( m \) is the slope (which is \( h'(3) = -108 \)) and \( b \) is the y-intercept.

You are given the point \( (3, -106) \). We can now substitute the slope \( m = -108 \) and the point \( (3, -106) \) into the equation of the tangent line to solve for \( b \):

\[
y = mx + b
\]
Substitute \( y = -106 \), \( x = 3 \), and \( m = -108 \):
\[
-106 = -108(3) + b
\]
\[
-106 = -324 + b
\]
\[
b = -106 + 324 = 218
\]

So the y-intercept \( b = 218 \).

### Final Equation of the Tangent Line:
\[
y = -108x + 218
\]

### Summary:
1. \( h'(3) = -108 \)
2. The equation of the tangent line is \( y = -108x + 218 \).

Would you like more details or clarifications? Here are 5 related questions:

1. How do you find the equation of a tangent line at any given point?
2. What are the general steps to compute a derivative of a polynomial function?
3. How do we interpret the derivative as the slope of a tangent line?
4. Why do we substitute the given point into the equation to solve for \( b \)?
5. How does the tangent line relate to the original curve at \( x = 3 \)?

**Tip:** Always verify the given point lies on the curve before proceeding with tangent line calculations.

If h(x) = 2 - 4x^3, find h'(3). Use this to find the equation of the tangent line to the curve y = 2 - 4x^3 at the point (3, -106). The equation of this tangent line can be written in the form y = mx + b, where m is the slope and b is the y-intercept.

Learn how to compute the derivative h'(x) of the function h(x) = 2 - 4x^3 and use it to find the equation of the tangent line at the point (3, -106). Suitable for calculus students, this problem covers key concepts of derivatives and tangent lines.

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