Linear functions and graphsIB MYP Maths Extended: Revision notes

A linear function has a graph that is a straight line. Its equation is $y=mx+c$, where $m$ is the gradient (steepness) and $c$ is the $y$-intercept (where the line crosses the $y$-axis). Example: $y=-2x+6$ has gradient $-2$ (it slopes downward) and $y$-intercept $6$, so it passes through $(0,6)$. A positive gradient slopes up from left to right and a negative gradient slopes down.

The gradient between $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2-y_1}{x_2-x_1}=\frac{\text{rise}}{\text{run}}.$ Example: for $A(1,3)$ and $B(5,11)$, $m=\frac{11-3}{5-1}=\frac84=2$. Subtract the coordinates in the same order on the top and bottom.

To plot a line, make a table of values, plot the points and join them with a ruler. To sketch a line quickly, find the intercepts. For $y=-2x+6$: when $x=0$, $y=6$, giving $(0,6)$. When $y=0$, $-2x+6=0$ so $x=3$, giving $(3,0)$. Mark both points and draw the line through them. Three points are better than two, because the third point checks for mistakes.

From the gradient and a point: put $m$ into $y=mx+c$, substitute the point and solve for $c$. From two points: find $m$ first, then do the same. Example: through $A(1,3)$ and $B(5,11)$. $m=2$, so $y=2x+c$. Using $(1,3)$: $3=2+c$, so $c=1$ and $y=2x+1$. To test whether a point lies on a line, substitute its $x$-value. For $(10,20)$ we get $y=2(10)+1=21\neq20$, so it is not on the line.

In a real situation the gradient is the rate of change, how much the output changes for every 1 unit of input. The $y$-intercept is the starting value (when $x=0$). Example: a pool fills so that $V=75t$. The gradient $75$ means 75 litres per minute, and the intercept $0$ means the pool starts empty. Always include units: say '75 litres per minute', not just '75'. To solve problems, set the equation equal to a given value, e.g. $75t=3000$ gives $t=40$ minutes.