Graphing Functions - SAT Math

$0$. Calculate $b^2 - 4ac = 16 - 16 = 0$.

$x = 2, x = 3$. Factors as $(x-2)(x-3) = 0$, so $x = 2$ or $x = 3$.

$(-3, 4)$. Vertex form shows $(h,k) = (-3,4)$ directly.

$x = -1$. Use $x = -\frac{b}{2a} = -\frac{4}{2(2)} = -1$.

Roots are solutions; factors are $(x - r_1)(x - r_2)$. If roots are $r_1, r_2$, then factors are $(x-r_1)(x-r_2)$.

$x = 2$. Axis of symmetry is $x = -\frac{b}{2a} = -\frac{-4}{2(1)} = 2$.

Minimum: $5$. Since $a > 0$, parabola opens up with minimum at vertex.

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Standard formula for solving quadratic equations.

$x = 2$ or $x = -2$. Factor as difference of squares: $(x-2)(x+2) = 0$.

$x = \frac{1}{2}$. Perfect square: $(2x-1)^2 = 0$ gives $x = \frac{1}{2}$.

$x = 2$ or $x = 3$. Factor: $(x-2)(x-3) = 0$ gives roots.

$x = 3i$ or $x = -3i$. Take square root: $x^2 = -9$, so $x = \pm 3i$.

1. Highest power of variable is 2.

$b^2 - 4ac$. Expression under the square root that determines root types.

$ -\frac{b}{a} $. Vieta's formula for sum of roots.

Complex solutions. Negative discriminant means no real roots.

$x = 3$ or $x = -3$. Difference of squares: $(x-3)(x+3) = 0$.

$x = 2$ or $x = -2$. Divide by 4: $x^2 = 4$, so $x = \pm 2$.

$\frac{c}{a}$. Vieta's formula for product of roots.

Number and type of roots. Determines number of real roots and their nature.

Completing the square. Transforms equation to perfect square trinomial form.

Complex. Discriminant $4 - 20 = -16 < 0$ means complex roots.

$ax^2 + bx + c = 0$. General form with degree 2 polynomial set to zero.

$x = 3$. Perfect square: $(x-3)^2 = 0$ has repeated root.

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

$y = (x+3)^2$. Perfect square: $(x+3)^2 = x^2 + 6x + 9$.

Two distinct real roots. When $b^2 - 4ac > 0$, the parabola crosses the x-axis twice.

$(3, 0)$. Perfect square: $-(x-3)^2$ has vertex at $(3,0)$.

$x = -3$. Perfect square trinomial: $(x+3)^2 = 0$, so $x = -3$.

$x = -2$. Perfect square: $(x+2)^2 = 0$ gives $x = -2$.

Parabola opens downward. Negative coefficient makes parabola open downward.

$x = -1$. Perfect square: $(x+1)^2$, so axis is $x = -1$.

$x = 1$ or $x = -1$. Difference of squares: $(x-1)(x+1) = 0$.

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Derived by completing the square on the general quadratic form.

Factored form. Shows roots $p$ and $q$ directly as factors.

Parabola opens upward. Positive coefficient makes parabola open upward.

Vertex: $(3, 4)$. In vertex form, $(h,k)$ gives the vertex coordinates directly.

$x \neq 2$. 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.

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.

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

Vertex form. Form that shows vertex $(h,k)$ and transformations.

$y = (x-2)^2 + 2$. Complete the square: $(x-2)^2 = x^2 - 4x + 4$.

$(1, 0)$. Perfect square: $(x-1)^2$ has vertex at $(1,0)$.

$0$. Calculate $b^2 - 4ac = 16 - 16 = 0$.

$y = 2(x+2)^2 - 2$. Complete the square: factor out 2, then add/subtract 4.

$ax^2 + bx + c = 0$. General form where $a \neq 0$ defines any quadratic equation.

$x = -3$. Use formula $x = -\frac{b}{2a} = -\frac{6}{2(1)} = -3$.

$(4, -7)$. Vertex form shows $(h,k) = (4,-7)$ directly.

$y = a(x-h)^2 + k$. Shows vertex $ (h,k) $ and stretch factor $a$ explicitly.

A function that is not a straight line. Graph is not a straight line (degree > 1).

$(1, 3)$. Vertex form shows $(h,k) = (1,3)$ directly.

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Derived by completing the square on the general quadratic form.

$ax^2 + bx + c = 0$. General form with degree 2 polynomial set to zero.

$\frac{c}{a}$. Vieta's formula for product of roots.

$-\frac{b}{a}$. Vieta's formula for sum of roots.

$y = a(x-h)^2 + k$. Shows vertex $(h,k)$ and stretch factor $a$ explicitly.

$b^2 - 4ac$. Expression under the square root that determines root types.

$x = 2$ or $x = 3$. Factor: $(x-2)(x-3) = 0$ gives roots.

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

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

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

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 $r$.

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

$y = 5$. 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.

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

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

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

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

$y = c$. 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.