To determine the gradient $m$ of the tangent line to the curve at $x = 1$, we need to calculate the slope of the tangent, which is the derivative $f'(x)$ evaluated at $x = 1$.

Looking at the graph, the orange line represents the tangent to the curve $y = f(x)$ at $x = 1$. The gradient of this tangent can be found by using the rise-over-run formula, that is, the change in $y$-coordinates divided by the change in $x$-coordinates between two points on the tangent line.

From the graph:

• At $x = 0$, the tangent has a $y$-coordinate of approximately $y = -1$.
• At $x = 2$, the tangent has a $y$-coordinate of approximately $y = -7$.

The slope $m$ can be calculated as:

$m = \frac{\Delta y}{\Delta x} = \frac{-7 - (-1)}{2 - 0} = \frac{-6}{2} = -3$

Thus, the gradient $m$ of the tangent line is $m = -3$.

Would you like further details or have any questions?

Here are 5 related questions:

1. How do you interpret the geometric meaning of a negative slope in a tangent line?
2. What does the derivative of a function represent in terms of the graph?
3. How can you use limits to define the derivative at a point?
4. How would you find the equation of the tangent line at $x = 1$?
5. Can you find the second derivative $f''(x)$, and what would it tell you about the curve?

Tip: The tangent line to a curve at a point represents the best linear approximation to the curve at that point, and its slope is equal to the derivative of the function at that point.