Slope and Intercepts: Practice

Read the $y$-intercept and the slope off the two marked points. The line crosses the $y$-axis at $(0, -1)$, so $b = -1$, and from $(0, -1)$ to $(2, 3)$ it rises $4$ over a run of $2$.

$m = \frac{3 - (-1)}{2 - 0} = \frac{4}{2} = 2, \qquad y = 2x - 1$

Only $y = 2x - 1$ passes through both marked points.

Solve for $y$ by subtracting $x$ from both sides.

$y = -x + 5$

So the slope is $-1$ and the $y$-intercept is $(0, 5)$.

Solve for $y$ to expose the slope.

$y = -3x + 7 \;\Rightarrow\; m = -3$

So the slope is $-3$; the coefficient $3$ in standard form becomes $-3$ once $y$ is isolated.

Set $x = 0$ and solve for $y$.

$3(0) + 2y = 12, \qquad 2y = 12, \qquad y = 6$

So the $y$-intercept is $(0, 6)$; $(4, 0)$ is the $x$-intercept.

The point $(0, 3)$ is the $y$-intercept, so $b = 3$, and the slope is $m = -\tfrac{1}{2}$.

$y = -\tfrac{1}{2}x + 3$

Substitute $m$ and $b$ straight into $y = mx + b$.

The line crosses the $y$-axis at $(0, -2)$, so $b = -2$, and from $(0, -2)$ to $(1, 1)$ it rises $3$ over a run of $1$.

$m = \frac{1 - (-2)}{1 - 0} = \frac{3}{1} = 3, \qquad y = 3x - 2$

Only $y = 3x - 2$ passes through both marked points.

Isolate the $y$-term, then divide every term by $3$.

$3y = -6x + 9, \qquad y = -2x + 3$

Dividing by $3$ must reach every term, turning $-6x$ into $-2x$ and $9$ into $3$.

Set $y = 0$ and solve for $x$.

$2x - 3(0) = 12, \qquad 2x = 12, \qquad x = 6$

So the $x$-intercept is $(6, 0)$; $(0, -4)$ is the $y$-intercept.

Slope-intercept form needs a defined slope $m$. A vertical line has an undefined slope, so it has no such form.

$x = 4 \;\Rightarrow\; \text{vertical, undefined slope}$

The horizontal line $y = 4$ is fine as $y = 0x + 4$, but $x = 4$ cannot be written as $y = mx + b$.

With no constant, $b = 0$, so the line passes through the origin.

$y = 5x = 5x + 0 \;\Rightarrow\; (0, 0)$

So the $y$-intercept is $(0, 0)$.

Distribute the slope, then subtract $1$ from both sides.

$y + 1 = -2x + 6, \qquad y = -2x + 5$

So the slope is $-2$ and the $y$-intercept is $(0, 5)$.

Set $y = 0$ and solve for $x$.

$0 = -3x + 9, \qquad 3x = 9, \qquad x = 3$

So the $x$-intercept is $(3, 0)$; $(0, 9)$ is the $y$-intercept.