We are given that the tangent line to the function $y = f(x)$ at the point $(-5, 4)$ passes through another point $(11, 11)$. We need to find $f(-5)$ and $f'(-5)$.

Step 1: Find $f(-5)$

The point $(-5, 4)$ is explicitly stated to be on the curve, meaning $f(-5) = 4$.

Step 2: Find $f'(-5)$

The derivative $f'(-5)$ represents the slope of the tangent line at the point $(-5, 4)$. Since the tangent line passes through both $(-5, 4)$ and $(11, 11)$, we can calculate the slope of the line passing through these two points.

The slope $m$ of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$

Substituting the points $(-5, 4)$ and $(11, 11)$: $m = \frac{11 - 4}{11 - (-5)} = \frac{7}{16}$

Therefore, the slope of the tangent line at $(-5, 4)$ is $f'(-5) = \frac{7}{16}$.

Final Answer:

• $f(-5) = 4$
• $f'(-5) = \frac{7}{16}$

Would you like more details or have any questions?

Here are some related questions:

1. How do we interpret the slope of a tangent line in a function graph?
2. What is the geometric meaning of a derivative at a point?
3. How do we use the point-slope form of a line equation?
4. Can the slope of a tangent line be used to estimate nearby values of the function?
5. How would this process change if the function was provided explicitly?

Tip: The slope of the tangent line at a point on a curve gives the rate of change of the function at that point.