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Let's go through the problem and identify what needs to be corrected for part (c). The problem involves the function \( f(x) = x^2 - x - 30 \) and requests different tasks related to secant and tangent lines.

### (a) Finding the equation of the secant line:
The secant line passes through the points \((-4, -10)\) and \((6, 0)\). The equation of the secant line is correctly calculated as:

\[
y = x - 6
\]

### (b) Using the Mean Value Theorem:
The Mean Value Theorem guarantees a point \( c \) in the interval \((-4, 6)\) such that the slope of the tangent line at \( c \) is equal to the slope of the secant line. Since the slope of the secant line is \( 1 \), we set the derivative of \( f(x) \) equal to \( 1 \).

\[
f'(x) = 2x - 1
\]

Setting \( f'(c) = 1 \):

\[
2c - 1 = 1 \implies c = 1
\]

### (c) Finding the equation of the tangent line through \( c \):
To find the tangent line at \( c = 1 \), we need to calculate the point on the curve and the slope of the tangent line at \( x = 1 \).

1. **Find the slope of the tangent line** at \( x = 1 \). We already know from part (b) that the slope at \( c = 1 \) is \( 1 \).

2. **Find the point on the curve** at \( x = 1 \):
\[
f(1) = 1^2 - 1 - 30 = -30
\]

So, the point on the curve is \( (1, -30) \).

3. **Write the equation of the tangent line**. Using the point-slope form of a line:
\[
y - y_1 = m(x - x_1)
\]
where \( m = 1 \), \( x_1 = 1 \), and \( y_1 = -30 \):

\[
y - (-30) = 1(x - 1)
\]
\[
y + 30 = x - 1
\]
\[
y = x - 31
\]

### Conclusion:
The correct equation of the tangent line through \( c = 1 \) is:

\[
y = x - 31
\]

Would you like further details or have any questions about the process?

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Here are 5 related questions to further your understanding:
1. How is the Mean Value Theorem used to find the point \( c \)?
2. Why is the slope of the secant line equal to the slope of the tangent line at \( c \)?
3. What is the geometric meaning of a secant line versus a tangent line?
4. How would you find the equation of the tangent line for a different function?
5. Can a function have more than one point where the Mean Value Theorem applies?

**Tip**: When finding the tangent line, always verify both the slope and the point on the curve to ensure the accuracy of your result.

Consider the function f(x) = x^2 - x - 30. (a) Find the equation of the secant line joining the points (-4, -10) and (6, 0). (b) Use the mean value theorem to determine a point c in the interval (-4, 6) such that the tangent line at c is parallel to the secant line. (c) Find the equation of the tangent line through c.

Learn how to use the mean value theorem to find the equation of secant and tangent lines for the quadratic function f(x) = x^2 - x - 30. This problem involves using calculus concepts such as derivatives and the mean value theorem.

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