Please forgive me, I forgot when first semester ended and second semester started, so some of the topics overlap... Alex is sorry
Slope: ($m$) $$ \large \frac{\Delta y}{\Delta x}$$
To find a perpendicular slope, take the negative reciprocal of the slope
e.g. the negative reciprocal of $8$ is $-\frac{1}{8}$
Midpoint: $(x_m, y_m)$
$$\large x_m = \frac{x_1 + x_2}{2}$$
$$\large y_m = \frac{y_1 + y_2}{2}$$Just average the two x coordinates and the two y coordinates
The Perpendicular Bisector will pass through the midpoint of the segment, and have a slope perpendicular to the segment
e.g. the perpendicular bisector of a segment with endpoints $(7,3)$ and $(-1,4)$ will pass through the point $(3,\frac{7}{2})$ with a slope of $8$ because the slope of the segment is equal to $-\frac{1}{8}$
$$\large\sqrt[n]{x} \times \sqrt[n]{y} = \sqrt[n]{xy}$$
$$\large\sqrt[n]{x} \div \sqrt[n]{y} = \sqrt[n]{\frac{x}{y}}$$
| Point | |
|---|---|
| x-intercept(s) | Solve the equation for $y$ (factoring, quadratic formula) |
| y-intercept | $c$ |
| Axis of Symmetry | $x = -\frac{b}{2a}$ |
$\leq or \geq\, \rightarrow [\, ]$
$< or >\, \rightarrow (\, )$
$x > -3$
$x < -7$
$x \leq -1$
$x \geq -9$
$[-9, -7)\cup(-3, -1]$
This means that $x$ can be in between $-9$ and $-7$, and can be $-9$ but can't be $-7$. $x$ is also in between $-3$ and $-1$, and can be $-1$, but can't be $-3$