We are asked to find the gradient (slope) of a straight line that passes through the point $(3, 6)$ and has a $y$-intercept of 3.

We know that the equation of a straight line is generally given by:

$y = mx + c$

where:

• $m$ is the gradient (slope),
• $c$ is the $y$-intercept.

In this problem, we are given:

• The $y$-intercept $c = 3$,
• The point $(x_1, y_1) = (3, 6)$ lies on the line.

Substitute these values into the equation:

$6 = m(3) + 3$

Solve for $m$ (the gradient):

$6 = 3m + 3$

$6 - 3 = 3m$

$3 = 3m$

$m = 1$

Thus, the gradient of the line is $m = 1$.

Would you like further clarification or details?

Here are 5 related questions to deepen understanding:

1. How do we derive the equation of a line when two points are given?
2. What does it mean if the gradient is negative?
3. How can you find the equation of a line if only the gradient and one point are given?
4. How do horizontal and vertical lines differ in terms of gradient?
5. Can you explain how to determine the intersection of two lines given their equations?

Tip: The gradient describes how steep a line is. A gradient of $1$ means that for every unit increase in $x$, $y$ also increases by 1.