Slope and Intercepts: Practice

The $y$-intercept is the constant $b$, written as the point $(0, b)$.

$y = 4x + 5 \;\Rightarrow\; b = 5, \;(0, 5)$

So the line crosses the $y$-axis at $(0, 5)$.

The $y$-intercept is the constant term with its sign, as the point $(0, b)$.

$y = x - 7 \;\Rightarrow\; b = -7, \;(0, -7)$

So the $y$-intercept is $(0, -7)$, not $(0, 7)$.

Slope-intercept form solves for $y$ and displays the slope $m$ and the intercept $b$.

$y = mx + b$

The form $Ax + By = C$ is standard form, and $y - y_1 = m(x - x_1)$ is point-slope form.

Put the equation in $y = mx + b$ order first, then read the coefficient of $x$.

$y = 5 - 2x = -2x + 5 \;\Rightarrow\; m = -2$

So the slope is $-2$, not the leading $5$.

The slope $3$ is the coefficient of $x$, and the intercept $(0, -2)$ makes $b = -2$.

$y = 3x - 2$

So the equation is $y = 3x - 2$; $y = -2x + 3$ swaps the two numbers.

A horizontal line never rises, so its slope is $0$. Written fully, $y = 7$ is $y = 0x + 7$.

$y = 7 \;\Rightarrow\; m = 0$

So the slope is $0$; a vertical line, not this one, has an undefined slope.

The slope $-1$ is the coefficient of $x$, and the intercept $(0, 6)$ gives $b = 6$.

$y = -x + 6$

So the equation is $y = -x + 6$.

The $y$-intercept is the constant term, as the point $(0, b)$.

$y = -\tfrac{2}{3}x + 4 \;\Rightarrow\; b = 4, \;(0, 4)$

The $-\tfrac{2}{3}$ is the slope, not the intercept.

The slope is the coefficient of $x$.

$y = \tfrac{1}{2}x - 5 \;\Rightarrow\; m = \tfrac{1}{2}$

So the slope is $\tfrac{1}{2}$; the $-5$ is the intercept.

A line through the origin has $y$-intercept $(0, 0)$, so $b = 0$ and the equation is $y = mx$.

$y = 4x = 4x + 0 \;\Rightarrow\; (0, 0)$

So $y = 4x$ passes through the origin; $y = 4x + 1$ crosses the $y$-axis at $(0, 1)$ instead.

A horizontal line fixes $y$ at a constant, the form $y = c$.

$y = -2 \;\Rightarrow\; \text{height } -2 \text{ at every } x$

So $y = -2$ is horizontal; $x = -2$ is vertical, and the others are slanted.

The $y$-intercept is the constant term with its sign, as $(0, b)$.

$y = -5x + 2 \;\Rightarrow\; b = 2, \;(0, 2)$

The $-5$ is the slope, not the intercept.