Solving Nonlinear Functions - SAT Math

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.

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.

$0$. 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 $x = -2$.

$y = a(x-h)^2 + k$. 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.

$ax^2 + bx + c = 0$. 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.

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