The tangent line to y=f(x)  at (-3, 3) passes through the point (10,-7). Compute the following. 1. f(-3)= and 2. f'(-3)

To solve this problem, we need to compute the values of \( f(-3) \) and \( f'(-3) \). Here’s how we proceed:

### 1. \( f(-3) \):
Since the point \((-3, 3)\) is on the curve of \( y = f(x) \), this tells us that:
\[
f(-3) = 3
\]

### 2. \( f'(-3) \):
The derivative \( f'(-3) \) represents the slope of the tangent line to the curve \( y = f(x) \) at the point \((-3, 3)\). We are given that this tangent line passes through the point \((10, -7)\).

We can find the slope of the tangent line by using the slope formula:
\[
\text{slope of the line} = \frac{y_2 - y_1}{x_2 - x_1}
\]
Substituting the points \((-3, 3)\) and \((10, -7)\):
\[
\text{slope of the line} = \frac{-7 - 3}{10 - (-3)} = \frac{-7 - 3}{10 + 3} = \frac{-10}{13}
\]
Thus, the slope of the tangent line (and therefore the value of \( f'(-3) \)) is:
\[
f'(-3) = -\frac{10}{13}
\]

### Final Answers:
1. \( f(-3) = 3 \)
2. \( f'(-3) = -\frac{10}{13} \)

Would you like more details or have any questions?

Here are 5 related questions:
1. How do we find the equation of the tangent line at a point?
2. Can a tangent line pass through more than one point on the curve?
3. What does the derivative \( f'(x) \) tell us about the behavior of the function?
4. How do we find the slope of a line given two points?
5. What is the geometric interpretation of \( f'(-3) \)?

**Tip:** The derivative at a point gives the slope of the tangent line at that specific point.

The tangent line to y=f(x) at (-3, 3) passes through the point (10,-7). Compute the following. 1. f(-3)= and 2. f'(-3)

Learn how to find the value of f(-3) and the derivative f'(-3) for a function where the tangent line at (-3, 3) passes through the point (10, -7). This problem involves the slope of a line and basic calculus differentiation, suitable for Grades 11-12 or early college students.

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