Slope and Intercepts: Practice

Solve for $y$ to read off the slope.

$-2y = -5x + 8, \qquad y = \tfrac{5}{2}x - 4$

So the slope is $\tfrac{5}{2}$; the $-4$ is the $y$-intercept.

The $y$-intercept gives $b = -2$, and the slope comes from the two intercept points.

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

So the equation is $y = \tfrac{1}{2}x - 2$.

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

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

Only $y = \tfrac{2}{3}x - 1$ passes through both marked points.

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

$3x - 4(0) = 18, \qquad 3x = 18, \qquad x = 6$

So the $x$-intercept is $(6, 0)$; setting $x = 0$ would give the $y$-intercept $\left(0, -\tfrac{9}{2}\right)$.

Write $y = -\tfrac{3}{5}x + b$ and use the point $(10, 0)$ to find $b$.

$0 = -\tfrac{3}{5}(10) + b, \qquad 0 = -6 + b, \qquad b = 6$

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

Steepness is measured by the size of the slope, its absolute value, ignoring the sign.

$|-4| = 4 > |3| = 3 > |1| = 1 > \left|\tfrac{1}{2}\right| = \tfrac{1}{2}$

So $y = -4x + 2$ is the steepest, even though its slope is negative.

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

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

Both marked points sit below the $x$-axis, but the slope is positive because the line climbs; only $y = 2x - 3$ fits.

Through the origin means $b = 0$, so the equation is $y = mx$ with $m = -\tfrac{5}{2}$.

$y = -\tfrac{5}{2}x$

The choice $y = -\tfrac{5}{2}x + 1$ misses the origin, and $y = \tfrac{5}{2}x$ has the wrong sign.

Compare the slopes and intercepts directly from each equation.

$m_A = m_B = 3, \qquad b_A = 2 \ne b_B = -4$

Both have slope $3$, so neither is steeper, but they cross the $y$-axis at different points, $(0, 2)$ and $(0, -4)$.

Isolate the $y$-term, then divide every term by $5$.

$5y = -2x, \qquad y = -\tfrac{2}{5}x$

The constant is $0$, so the line passes through the origin with slope $-\tfrac{2}{5}$.

The value $b$ is the $y$-intercept, so $b = 0$ means the line crosses the $y$-axis at $(0, 0)$.

$y = mx + 0 = mx \;\Rightarrow\; (0, 0) \text{ is on the line}$

So the line passes through the origin; its slope $m$ can still be any value.

Solve for $y$, or set $x = 0$ directly.

$-y = -7x + 3, \qquad y = 7x - 3 \;\Rightarrow\; (0, -3)$

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