Once an equation is written in the form $y=mx+c$, the gradient and intercept can be read off directly — no drawing required:

State the gradient and $y$-intercept of $y=6x+3$: comparing directly, $m=6$ and $c=3$.

Lines may also appear as $x=k$ (a vertical line through $x=k$) or $y=k$ (a horizontal line through $y=k$) — don't mix these two up. Neither of these special forms has a "$c$" in the usual sense: $y=k$ has a gradient of $0$, while $x=k$ has no defined gradient at all.

1. Read the $y$-intercept directly from where the line crosses the $y$-axis — this is $c$.
2. Find the gradient $m$ from two clear points on the line (rise over run).
3. Substitute both values into $y=mx+c$.

A graph passes through $(0,-2)$ and $(4,6)$:

1. $c=-2$ (read directly, since $x=0$ is one of the points).
2. $m = \dfrac{6-(-2)}{4-0} = \dfrac{8}{4} = 2$.
3. Equation: $y = 2x-2$.

Substitute the gradient into $y=mx+c$, then substitute the known point's coordinates to solve for $c$.

A line has gradient $3$ and passes through $(2,11)$:

1. $y=3x+c$.
2. Substitute the point: $11=3(2)+c \implies c=5$.
3. Equation: $y = 3x+5$.

A horizontal line has the same $y$-value at every single point on it, so its equation is simply $y=(\text{that value})$ — it has no $x$ term at all. Likewise, a vertical line has the same $x$-value throughout, giving an equation of the form $x=(\text{that value})$, with no $y$ term.

A line passes through $(3,-2)$ and $(3,5)$: since both points share $x=3$, this is a vertical line, with equation $x=3$ — not an equation involving $y$ at all.

A straight-line graph has gradient $-4$ and passes through $(3,-1)$. Find its equation, giving your answer in the form $y=mx+c$.

Solution

1. Substitute $m=-4$: $y=-4x+c$.
2. Substitute the point $(3,-1)$: $-1 = -4(3)+c \implies -1=-12+c \implies c=11$.
3. Final equation:

Always leave the final equation fully simplified — Cambridge examiners expect $y=-4x+11$, not an unsimplified equivalent like $y=\frac{-12}{3}x+11$.

If an equation is given as $x=k$ or in a rearranged form rather than directly as $y=mx+c$, convert it first before trying to compare gradients or intercepts — mixing up a vertical line's equation with an ordinary sloped line is a common Core-tier error, since $x=k$ has no $y$-intercept or gradient in the usual sense at all.