The formula y = mx + b

The formula of a linear function tells at once what its line looks like:

• $\displaystyle m$ – the slope of the line.
• $\displaystyle b$ – where the line meets the $\displaystyle y$-axis: at the point $\displaystyle (0,b)$.

• Substitute $\displaystyle {x}={0}$: $\displaystyle {y}={m\cdot0+b}={b}$. So the line passes through the point $\displaystyle (0,b)$, on the $\displaystyle y$-axis.
• When $\displaystyle x$ grows by 1, $\displaystyle mx$ grows by $\displaystyle m$, and $\displaystyle b$ does not change. So $\displaystyle y$ grows by $\displaystyle m$ – and that is the slope.

• $\displaystyle {y}={2x-3}$: $\displaystyle {m}={2}$, $\displaystyle {b}={-3}$.
• $\displaystyle {y}={4-3x}$: $\displaystyle {m}={-3}$, $\displaystyle {b}={4}$.
• $\displaystyle {y}={x+1}$: $\displaystyle {m}={1}$, since $\displaystyle {x}={1\cdot x}$.
• $\displaystyle {y}={-x}$: $\displaystyle {m}={-1}$, $\displaystyle {b}={0}$.
• $\displaystyle {y}={\frac{x}{2}+5}$: $\displaystyle {m}={\frac{1}{2}}$, since $\displaystyle {\frac{x}{2}}={\frac{1}{2}x}$.

Two points fix a line, and the formula $\displaystyle {y}={mx+b}$ gives two of them without a table.
We draw $\displaystyle {y}={2x-3}$, where $\displaystyle {m}={2}$ and $\displaystyle {b}={-3}$:

• We mark $\displaystyle (0,-3)$ on the $\displaystyle y$-axis.
• The slope is 2: from there one step right and two up – the point $\displaystyle (1,-1)$.
• We draw a line through the two points.

When the slope is a fraction, it is handy to move right by the denominator.
In $\displaystyle {y}={-\frac{1}{2}x+4}$:

• We mark $\displaystyle (0,4)$.
• The slope is $\displaystyle -\frac{1}{2}$: two steps right, and one down – the point $\displaystyle (2,3)$.

• $\displaystyle {y}={3x}$: $\displaystyle {b}={0}$. The line passes through the origin, $\displaystyle (0,0)$.
• $\displaystyle {y}={5}$: $\displaystyle {m}={0}$. The line is horizontal, and meets the $\displaystyle y$-axis at $\displaystyle (0,5)$.

On the $\displaystyle x$-axis every point has $\displaystyle {y}={0}$. So we substitute $\displaystyle {y}={0}$ in the formula and solve an equation.
In $\displaystyle {y}={2x-3}$:

• $\displaystyle {0}={2x-3}$.
• $\displaystyle {2x}={3}$.
• $\displaystyle {x}={1.5}$: the line meets the $\displaystyle x$-axis at the point $\displaystyle (1.5,0)$.

For each line, write its slope and where it meets the $\displaystyle y$-axis.

• $\displaystyle {y}={3x+1}$
• $\displaystyle {y}={-2x+5}$
• $\displaystyle {y}={\frac{1}{4}x-2}$
• $\displaystyle {y}={6-x}$
• $\displaystyle {y}={-7}$
• $\displaystyle {y}={\frac{2}{3}x}$

Find where each line meets the $\displaystyle x$-axis.
In part c, write the equation yourself.

Work out where the line $\displaystyle {y}={-2x+4}$ meets each of the axes.
Mark the points.
Then show the line, and check that it passes through them.

For each of the lines $\displaystyle {y}={3x-2}$ and $\displaystyle {y}={-\frac{1}{2}x+3}$, mark two points without a table: where it meets the $\displaystyle y$-axis, and one more point by the slope.
Then show the lines, and check.

Here are some lines:

• $\displaystyle {y}={2x-5}$
• $\displaystyle {y}={x+1}$
• $\displaystyle {y}={3+2x}$
• $\displaystyle {y}={-2x+1}$
• $\displaystyle {y}={\frac{1}{2}x}$

Which of them are parallel to the line $\displaystyle {y}={2x+1}$?
Which of them meet the $\displaystyle y$-axis at the same point as $\displaystyle {y}={2x+1}$?