Quadratic Equations - SAT Math

$ax^2 + bx + c = 0$. General form where $a \neq 0$ determines parabola shape.

$b^2 - 4ac$. The expression under the square root in the quadratic formula.

$(x - 3)(x + 3)$. Difference of squares: $a^2 - b^2 = (a-b)(a+b)$.

$b^2 = 4ac$. When discriminant equals zero, the quadratic is a perfect square.

$x = 2, x = -4$. Factor as $(x-2)(x+4) = 0$ or use the quadratic formula.

$x = \frac{-b}{2a}$. The $x$-coordinate of the vertex, found by calculus or completing the square.

$x = -2, x = 2$. Set $y = 0$: $x^2 - 4 = 0$, so $x^2 = 4$.

If $ab = 0$, then $a = 0$ or $b = 0$. If a product equals zero, at least one factor must be zero.

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

$ax^2 + bx + c = 0$. The general form where $a \neq 0$ defines a parabola.

Narrower parabola. Larger $|a|$ values compress the parabola horizontally.

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

$(x + 4)^2$. Perfect square trinomial with $a = x$, $b = 4$.

$x^2 + 4x + 4$. Apply the perfect square formula $(a+b)^2 = a^2 + 2ab + b^2$.

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

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Solves any quadratic by substituting coefficients $a$, $b$, and $c$.

$\frac{c}{a}$. By Vieta's formulas relating coefficients to roots.

$y = (x-2)^2$. This is $(x-2)^2$ expanded, so vertex form shows vertex $(2,0)$.

$x = \frac{-b}{2a}$. Derived from completing the square or using calculus.

$y = 2x^2 - 12x + 22$. Expand $(x-3)^2 = x^2 - 6x + 9$, then distribute and combine.

A downward-opening parabola. Negative coefficient of $x^2$ makes the parabola open downward.

$y = (x+3)^2 - 1$. Complete the square: $(x+3)^2 = x^2 + 6x + 9$, so subtract 1.

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

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

$x = 2, x = -2$. Divide by 3: $x^2 - 4 = 0$, so $x = \pm 2$.

Vertex: $(-2, 0)$. This is $(x+2)^2$, so vertex is at $x = -2$, $y = 0$.

$x = 2, x = 3$. Factor as $(x-2)(x-3) = 0$ to find roots.

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

The coefficient $a$. Positive $a$ opens up, negative $a$ opens down.

The quadratic formula. Works for any quadratic, unlike factoring or square roots.

Vertex: $(1, 3)$. In vertex form $y = a(x-h)^2 + k$, vertex is $(h,k)$.

$-\frac{b}{a}$. By Vieta's formulas relating coefficients to roots.

If $b^2 - 4ac > 0$, 2 real roots; $=0$, 1 real root; $<0$, no real roots. The discriminant determines whether roots are real or complex.

Intercept: $(0, 1)$. Set $x = 0$ to find where the parabola crosses the $y$-axis.

Vertex: $(2, 1)$. Use $x = -\frac{b}{2a} = 2$, then $y = -(4) + 8 - 3 = 1$.

Minimum: $0$. This is $(x-3)^2$, so minimum occurs at vertex $(3,0)$.

$x = 2, x = 3$. Factor as $(x-2)(x-3) = 0$ or use the quadratic formula.

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

Vertex: $(2, -1)$. Use $x = \frac{4}{2} = 2$, then $y = 4 - 8 + 3 = -1$.

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

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Solves any quadratic by substituting coefficients $a$, $b$, and $c$.

$ax^2 + bx + c = 0$. General form where $a \neq 0$ determines parabola shape.

$x = \frac{-b}{2a}$. The $x$-coordinate of the vertex, found by calculus or completing the square.

Vertex: $(-2, 0)$. This is $(x+2)^2$, so vertex is at $x = -2$, $y = 0$.

The quadratic formula. Works for any quadratic, unlike factoring or square roots.

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

Opens upwards. Positive coefficient of $x^2$ means parabola opens upward.

$x = -2, x = 2$. Set $y = 0$: $x^2 - 4 = 0$, so $x^2 = 4$.

Narrower parabola. Larger $|a|$ values compress the parabola horizontally.

$(x + 4)^2$. Perfect square trinomial with $a = x$, $b = 4$.

$x = 2, x = -2$. Divide by 3: $x^2 - 4 = 0$, so $x = \pm 2$.

Intercept: $(0, 1)$. Set $x = 0$ to find where the parabola crosses the $y$-axis.

$y = (x-2)^2$. This is $(x-2)^2$ expanded, so vertex form shows vertex $(2,0)$.

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

Minimum: $0$. This is $(x-3)^2$, so minimum occurs at vertex $(3,0)$.

$x = 2, x = -4$. Factor as $(x-2)(x+4) = 0$ or use the quadratic formula.

$b^2 = 4ac$. When discriminant equals zero, the quadratic is a perfect square.

$y = (x+3)^2 - 1$. Complete the square: $(x+3)^2 = x^2 + 6x + 9$, so subtract 1.

$x^2 + 4x + 4$. Apply the perfect square formula $(a+b)^2 = a^2 + 2ab + b^2$.

If $ab = 0$, then $a = 0$ or $b = 0$. If a product equals zero, at least one factor must be zero.

$-\frac{b}{a}$. By Vieta's formulas relating coefficients to roots.

A downward-opening parabola. Negative coefficient of $x^2$ makes the parabola open downward.

$(x - 3)(x + 3)$. Difference of squares: $a^2 - b^2 = (a-b)(a+b)$.

Vertex: $(2, 1)$. Use $x = -\frac{b}{2a} = 2$, then $y = -(4) + 8 - 3 = 1$.

$x = 2, x = 3$. Factor as $(x-2)(x-3) = 0$ or use the quadratic formula.

$\frac{c}{a}$. By Vieta's formulas relating coefficients to roots.

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

$ax^2 + bx + c = 0$. The general form where $a \neq 0$ defines a parabola.

If $b^2 - 4ac > 0$, 2 real roots; $=0$, 1 real root; $<0$, no real roots. The discriminant determines whether roots are real or complex.

$b^2 - 4ac$. The expression under the square root in the quadratic formula.

Vertex: $(1, 3)$. In vertex form $y = a(x-h)^2 + k$, vertex is $(h,k)$.