Graphs - SAT Math

Solve $Ax + By = C$ for $y$. Isolate $y$ by dividing by coefficient of $y$.

$y = mx + b$. Standard form where $m$ is slope and $b$ is y-intercept.

$y = a(x - h)^2 + k$. Shows vertex $(h, k)$ and vertical shift $k$.

The rate of change or steepness of the line. Measures how much $y$ changes per unit increase in $x$.

$x = -\frac{b}{2a}$. Vertical line through the vertex of the parabola.

$(2, 0)$. Set $y = 0$ and solve for $x$.

$(x - h)^2 + (y - k)^2 = r^2$. Standard circle equation with center $(h, k)$ and radius $r$.

$x^2 + y^2 + Dx + Ey + F = 0$. Expanded form of circle equation.

(3, 2). Vertex form $(x - h)^2 + k$ has vertex at $(h, k)$.

$\frac{1}{3}$. Coefficient of $(x - 3)$ in point-slope form.

$y = 3x + 1$. Same slope, substitute point to find $b$.

-4. Parallel lines have identical slopes.

$y = 4x$. Negative reciprocal slope through given point.

Downward. Negative coefficient of $x^2$ opens downward.

$ax^2 + bx + c = 0$. Standard form with highest power of 2.

1. The constant term when $x = 0$.

$\frac{3}{4}$. Rearrange to $y = \frac{3}{4}x - 3$ to identify slope.

$-\frac{1}{2}$. Negative reciprocal of the original slope.

Point-slope form. Uses a known point $(x_1, y_1)$ and slope $m$.

$x = a$. All points have the same $x$-coordinate.

1. Set $y = 0$ and solve: $0 = 2x - 8$, so $x = 4$.

(1, 5). Substituting: $y = 2(1) + 3 = 5$, so point $(1, 5)$ satisfies the equation.

$\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$. Average of the coordinates of the endpoints.

$m = \frac{y_2 - y_1}{x_2 - x_1}$. Change in $y$ over change in $x$ between two points.

$Ax + By = C$. Linear equation with integer coefficients $A$, $B$, and $C$.

$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Pythagorean theorem applied to coordinate plane differences.

$y = b$. All points have the same $y$-coordinate.

$(3, 4)$. Average the x-coordinates and y-coordinates: $\frac{1+5}{2}, \frac{2+6}{2}$.

$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Pythagorean theorem applied to coordinate plane.

$(3, -4)$. Values of $(h, k)$ from standard form.

$y = mx + b$. Shows slope $m$ and $y$-intercept $b$ directly.

$y = mx + b$. Standard form where $m$ is slope and $b$ is y-intercept.

$y = ax^2 + bx + c$. Quadratic with degree 2 polynomial.

$Ax + By = C$. Linear equation with integer coefficients and no fractions.

1. Rearrange to $y = mx + b$ form first.

x-intercepts: $3, -2$. Set $y = 0$ to find where parabola crosses $x$-axis.

$(x - h)^2 + (y - k)^2 = r^2$. Distance from center to any point on circle equals radius.

$(0, 2)$. Set $x = 0$ and solve for $y$.

Vertical. Constant $x$-value creates vertical line.

$(x - h)^2 + (y - k)^2 = r^2$. Distance from center to any point on circle.

1. Horizontal line has zero rise over run.

Each input has exactly one output. Vertical line test ensures unique outputs.

$4$. Square root of the constant term.

Right 1 unit, up 2 units. Vertex form shows horizontal and vertical shifts.

1. Solve for $x$ when $y = 0$.

x-intercept: 6, y-intercept: -3. Set each variable to zero alternately.

Parabola. Quadratic function creates U-shaped curve.

1. Square root of the constant gives radius.

1. Set $x = 0$: $5y = 20$, so $y = 4$.

-5. Coefficient of $x$ in slope-intercept form.

$(1, 3)$. Solve the system by setting equations equal.

$(2, -3)$. Vertex form shows $(h, k)$ as the turning point.

$(3, 4)$. Average the x-coordinates and y-coordinates: $\frac{1+5}{2}, \frac{2+6}{2}$.

Point-slope form. Uses a known point $(x_1, y_1)$ and slope $m$.

1. Set $y = 0$ and solve: $0 = 2x - 8$, so $x = 4$.

$\frac{3}{4}$. Rearrange to $y = \frac{3}{4}x - 3$ to identify slope.

1. Set $x = 0$: $5y = 20$, so $y = 4$.

$y = mx + b$. Standard form where $m$ is slope and $b$ is y-intercept.

$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Pythagorean theorem applied to coordinate plane differences.

Parabola. Quadratic function creates U-shaped curve.

$(x - h)^2 + (y - k)^2 = r^2$. Distance from center to any point on circle.

$y = mx + b$. Shows slope $m$ and $y$-intercept $b$ directly.

1. The constant term when $x = 0$.

$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Pythagorean theorem applied to coordinate plane.

1. Solve for $x$ when $y = 0$.

$ax^2 + bx + c = 0$. Standard form with highest power of 2.

$(2, -3)$. Vertex form shows $(h, k)$ as the turning point.

$y = a(x - h)^2 + k$. Shows vertex $(h, k)$ and vertical shift $k$.

$-\frac{1}{2}$. Negative reciprocal of the original slope.

-4. Parallel lines have identical slopes.

1. Square root of the constant term.

1. Rearrange to $y = mx + b$ form first.

1. Horizontal line has zero rise over run.

Vertical. Constant $x$-value creates vertical line.

$7$. Square root of the constant gives radius.

$y = ax^2 + bx + c$. Quadratic with degree 2 polynomial.

$\frac{1}{3}$. Coefficient of $(x - 3)$ in point-slope form.

$y = 4x$. Negative reciprocal slope through given point.

x-intercepts: $3, -2$. Set $y = 0$ to find where parabola crosses $x$-axis.

$(1, 3)$. Solve the system by setting equations equal.

Each input has exactly one output. Vertical line test ensures unique outputs.

x-intercept: 6, y-intercept: -3. Set each variable to zero alternately.

$x^2 + y^2 + Dx + Ey + F = 0$. Expanded form of circle equation.

$m = \frac{y_2 - y_1}{x_2 - x_1}$. Change in $y$ over change in $x$ between two points.

Right 1 unit, up 2 units. Vertex form shows horizontal and vertical shifts.

$(-2, -5)$. Vertex form identifies $(h, k)$ directly.

$Ax + By = C$. Linear equation with integer coefficients and no fractions.

$y = 3x + 1$. Same slope, substitute point to find $b$.

The rate of change or steepness of the line. Measures how much $y$ changes per unit increase in $x$.

-5. Coefficient of $x$ in slope-intercept form.

$\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$. Average of the coordinates of the endpoints.

$y = b$. All points have the same $y$-coordinate.

$(3, -4)$. Values of $(h, k)$ from standard form.

$x = -\frac{b}{2a}$. Vertical line through the vertex of the parabola.

$(2, 0)$. Set $y = 0$ and solve for $x$.

$x = a$. All points have the same $x$-coordinate.

Downward. Negative coefficient of $x^2$ opens downward.

$(0, 2)$. Set $x = 0$ and solve for $y$.

Solve $Ax + By = C$ for $y$. Isolate $y$ by dividing by coefficient of $y$.

(1, 5). Substituting: $y = 2(1) + 3 = 5$, so point $(1, 5)$ satisfies the equation.

(3, 2). Vertex form $(x - h)^2 + k$ has vertex at $(h, k)$.

$y = mx + b$. Standard form where $m$ is slope and $b$ is y-intercept.

$(x - h)^2 + (y - k)^2 = r^2$. Standard circle equation with center $(h, k)$ and radius $r$.