Honors Algebra II Unit 5 Review: Quadratic Functions and Complex Numbers

A quadratic function has degree 2, so its standard form is $f(x) = ax^2 + bx + c$ with $a \neq 0$. Its graph is a parabola that opens up when $a > 0$ (the vertex is a minimum) and down when $a < 0$ (the vertex is a maximum). The axis of symmetry is $x = -\frac{b}{2a}$, and the y-intercept is $(0, c)$.

In vertex form $f(x) = a(x - h)^2 + k$, the graph of $y = x^2$ is scaled vertically by $|a|$ and reflected across the x-axis if $a < 0$. It is then shifted horizontally by $h$ and vertically by $k$. Watch the sign: $(x + 3)^2$ means $h = -3$, a shift left.

You can solve $ax^2 + bx + c = 0$ by factoring, completing the square, or using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Graphing can show real solutions, but not nonreal solutions. The domain is all real numbers, and the range starts or stops at $k$.

Worked example. Take $f(x) = 2x^2 - 8x + 3$. The vertex x-value is $-\frac{-8}{2(2)} = 2$, and $f(2) = 8 - 16 + 3 = -5$, so the vertex is $(2, -5)$ and vertex form is $f(x) = 2(x - 2)^2 - 5$. Since $a = 2 > 0$, the range is $y \geq -5$. For the zeros, the discriminant is $64 - 24 = 40$, so there are two real roots: $x = \frac{8 \pm \sqrt{40}}{4} = 2 \pm \frac{\sqrt{10}}{2}$, about $3.58$ and $0.42$.

The imaginary unit $i$ is defined by $i^2 = -1$, so $\sqrt{-16} = 4i$. A complex number has the form $a + bi$, where $a$ is the real part and $b$ is the imaginary part (just the coefficient, not $bi$). Every real number is complex with $b = 0$.

To add or subtract, combine real parts and imaginary parts separately: $(3 + 2i) + (4 - 5i) = 7 - 3i$. To multiply, distribute and replace $i^2$ with $-1$: $(2 + 3i)(4 - i) = 8 - 2i + 12i - 3i^2 = 11 + 10i$. Powers of $i$ repeat every four steps ($i, -1, -i, 1$), so $i^{23} = i^3 = -i$ because 23 leaves remainder 3 when divided by 4.

The conjugate of $a + bi$ is $a - bi$, and their product $a^2 + b^2$ is real. To divide, multiply top and bottom by the denominator's conjugate. For example, $\frac{3 + 4i}{1 - 2i} = \frac{(3 + 4i)(1 + 2i)}{5} = \frac{-5 + 10i}{5} = -1 + 2i$.

Worked example. Solve $x^2 + 2x + 5 = 0$. The discriminant is $2^2 - 4(1)(5) = -16$, so expect two complex conjugate solutions. Then $x = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i$. The solutions $-1 + 2i$ and $-1 - 2i$ are conjugates. Nonreal solutions of a real-coefficient quadratic always occur in conjugate pairs.

A quadratic-linear system pairs a parabola with a line, and its solutions are their intersection points. For a nonvertical line, solve its equation for $y$ and substitute into the quadratic. Solve the resulting equation, then plug each $x$ back in to get $y$. For a vertical line $x = c$, substitute $c$ directly into the quadratic function to find $y$.

After substituting, the discriminant of that new quadratic tells you how many intersections there are. Positive means the line crosses the parabola twice, zero means the line is tangent (one point), and negative means they never meet, so there are no real solutions.

Worked example. Solve $y = x^2 - 3x + 2$ and $y = x - 1$. Set them equal to get $x^2 - 3x + 2 = x - 1$, which becomes $x^2 - 4x + 3 = 0$, or $(x - 1)(x - 3) = 0$. So $x = 1$ or $x = 3$, and the line gives $y = 0$ and $y = 2$. The solutions are $(1, 0)$ and $(3, 2)$.

For modeling, the vertex answers max and min questions. A ball thrown up at 48 ft/s from the ground has height $h(t) = -16t^2 + 48t$. The peak is at $t = -\frac{48}{2(-16)} = 1.5$ s, with $h(1.5) = 36$ ft, and it lands when $-16t(t - 3) = 0$, at $t = 3$ s. Always ask which solutions make sense in context, since a negative time gets thrown out.