Graphing Functions - SAT Math

All real numbers. Natural logarithm outputs all real values for positive inputs.

$f(x) = -x^3$. Reflection over y-axis changes $x$ to $-x$ in odd functions.

$y = 7$. When $x = 0$, $y = -3(0) + 7 = 7$.

$(3, 4)$. Vertex form directly shows vertex coordinates $(h, k)$.

$y = mx + b$. Standard linear form with slope $m$ and y-intercept $b$.

Hyperbola. Rational function with two branches in opposite quadrants.

$y = a \cdot b^x$, $b > 1$. Base $b > 1$ creates increasing exponential function.

$y = ax^2 + bx + c$. General quadratic form with leading coefficient $a$ and constant term $c$.

$y = a(x - h)^2 + k$. Shows vertex at $(h, k)$ with horizontal shifts and vertical shifts.

$y = -3$. Exponential functions approach horizontal asymptotes as $x \to -\infty$.

$x = -2$. Set $y = 0$ and solve: $0 = 3x + 6$, so $x = -2$.

$x = 3, x = -3$. Factor as $(x-3)(x+3) = 0$ to find roots.

$(2, 0)$. Perfect square trinomial $(x-2)^2$ has double root at $x = 2$.

$(3, 0)$. Complete the square: $(x-3)^2$ gives vertex at $(3, 0)$.

$y \geq 0$. Absolute value function always produces non-negative outputs.

$y = ab^x$. Base $b$ raised to variable power $x$ with coefficient $a$.

$m = -5$. Coefficient of $x$ in linear form $y = mx + b$.

$y \leq 4$. Parabola opens downward with maximum value $y = 4$.

$y = 7$. Y-intercept occurs when $x = 0$, giving $y = 7$.

$y = c$. Horizontal lines have zero slope and constant y-value.

Right 4, Up 2. $(x - 4)$ shifts right 4, $+2$ shifts up 2.

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

Undefined. Vertical lines have infinite slope, expressed as undefined.

$y = ax^2 + bx + c$, $a > 0$. When $a > 0$, the parabola opens upward with minimum vertex.

$x = -\frac{3}{2}$. Set $y = 0$ and solve: $0 = 2x + 3$, so $x = -\frac{3}{2}$.

$y = a(x - h)^2 + k$. Vertex form with $a > 0$ opens upward from vertex $(h, k)$.

$m = \frac{y_2 - y_1}{x_2 - x_1}$. Rise over run formula between two points.

Left 5. $(x + 5)$ represents horizontal shift left by 5 units.

$x \geq 3$. Square root requires non-negative input, so $x - 3 \geq 0$.

Reflection over x-axis. Negative coefficient flips parabola upside down.

$x \neq 2$. Denominator cannot equal zero, so $x - 2 \neq 0$.

$y \geq -5$. Parabola opens upward with minimum value at vertex $y = -5$.

$y = 5$. Y-intercept is the constant term when $x = 0$.

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

$y = a(x-h)^2 + k$. Standard vertex form with vertex at $(h,k)$ and vertical stretch $a$.

$x^2 + y^2 = r^2$. Circle equation with center $(0,0)$ and radius $r$.

Vertical stretch by 3. Coefficient greater than 1 stretches graph vertically.

$x \neq 3, x \neq -2$. Factor: $(x-3)(x+2) \neq 0$, so exclude both roots.

$(3, 0)$. Complete the square: $(x-3)^2$ gives vertex at $(3, 0)$.

$y = ax^2 + bx + c$. General quadratic form with leading coefficient $a$ and constant term $c$.

$(2, 0)$. Perfect square trinomial $(x-2)^2$ has double root at $x = 2$.

$y = c$. Horizontal lines have zero slope and constant y-value.

$x = 3, x = -3$. Factor as $(x-3)(x+3) = 0$ to find roots.

All real numbers. Natural logarithm outputs all real values for positive inputs.

Undefined. Vertical lines have infinite slope, expressed as undefined.

$m = -5$. Coefficient of $x$ in linear form $y = mx + b$.

$x \geq 4$. Square root requires non-negative argument: $x - 4 \geq 0$.

$f(x) = -x^3$. Reflection over y-axis changes $x$ to $-x$ in odd functions.

Right 4, Up 2. $(x - 4)$ shifts right 4, $+2$ shifts up 2.

Vertical stretch by 3. Coefficient greater than 1 stretches graph vertically.

Left 5. $(x + 5)$ represents horizontal shift left by 5 units.

$y = mx + b$. Standard linear form with slope $m$ and y-intercept $b$.

$y = 7$. Y-intercept occurs when $x = 0$, giving $y = 7$.

$y = -3$. Exponential functions approach horizontal asymptotes as $x \to -\infty$.

$y = a(x - h)^2 + k$. Shows vertex at $(h, k)$ with horizontal shifts and vertical shifts.

$x^2 + y^2 = r^2$. Circle equation with center $(0,0)$ and radius $r$.

$y \leq 4$. Parabola opens downward with maximum value $y = 4$.

$y = 5$. Y-intercept is the constant term when $x = 0$.

$x \neq 3, x \neq -2$. Factor: $(x-3)(x+2) \neq 0$, so exclude both roots.

$x = -2$. Set $y = 0$ and solve: $0 = 3x + 6$, so $x = -2$.

$(3, 4)$. Vertex form directly shows vertex coordinates $(h, k)$.

$y \geq -5$. Parabola opens upward with minimum value at vertex $y = -5$.

Reflection over x-axis. Negative coefficient flips parabola upside down.

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

$y = a(x - h)^2 + k$. Vertex form with $a > 0$ opens upward from vertex $(h, k)$.

$m = \frac{y_2 - y_1}{x_2 - x_1}$. Rise over run formula between two points.

$y = ab^x$. Base $b$ raised to variable power $x$ with coefficient $a$.

$x \neq 2$. Denominator cannot equal zero, so $x - 2 \neq 0$.