Quadratic Functions & Complex Numbers

Study Sheet

Quadratic Functions & Complex Numbers

Three forms, graphing, solving, completing the square, the quadratic formula, and complex numbers

Three Forms of a Quadratic

Concept
The same parabola, written three ways
xy

Every quadratic function can be written in three equivalent forms. Each form reveals something different at a glance.

  • Standard form: y=ax2+bx+cy=ax^2+bx+c  (reveals the yy-intercept cc)
  • Vertex form: y=a(x−h)2+ky=a(x-h)^2+k  (reveals the vertex (h,k)(h,k))
  • Factored form: y=a(x−r1)(x−r2)y=a(x-r_1)(x-r_2)  (reveals the xx-intercepts r1,r2r_1,r_2)

In all three, the number aa is the same: it controls direction and width.

The vertex is the turning point.

Example
One parabola in all three forms

Consider y=x2−2x−3y=x^2-2x-3 (standard). Its vertex is at x=−b2a=22=1x=-\dfrac{b}{2a}=\dfrac{2}{2}=1, giving y=1−2−3=−4y=1-2-3=-4, so

y=(x−1)2−4(vertex form, vertex (1,−4)).y=(x-1)^2-4 \quad\text{(vertex form, vertex }(1,-4)).

Factoring the standard form: x2−2x−3=(x+1)(x−3)x^2-2x-3=(x+1)(x-3), so

y=(x+1)(x−3)(factored form, roots x=−1, x=3).y=(x+1)(x-3)\quad\text{(factored form, roots }x=-1,\ x=3).
Tip

Tip: To go from standard to vertex form, complete the square. To go from standard to factored form, factor (or use the roots from the quadratic formula). To go back to standard from either, expand and combine like terms.

Reminder — The quadratic formula:  x=−b±b2−4ac2a(a≠0)x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad (a\neq 0)

Graphing Parabolas

Concept
Anatomy of a parabola
xy

For y=ax2+bx+cy=ax^2+bx+c:

  • Direction: opens up if a>0a>0 (has a minimum), down if a<0a<0 (has a maximum).
  • Axis of symmetry: the vertical line x=−b2ax=-\dfrac{b}{2a}.
  • Vertex: (−b2a, f ⁣(−b2a))\left(-\dfrac{b}{2a},\ f\!\left(-\dfrac{b}{2a}\right)\right); this is the min or max point.
  • yy-intercept: (0,c)(0,c).  xx-intercepts: solve ax2+bx+c=0ax^2+bx+c=0.

The vertex is the turning point.

Example
Reading a graph

For y=−x2+3y=-x^2+3: a=−1<0a=-1<0 so it opens down with a maximum. The axis of symmetry is x=0x=0 and the vertex is (0,3)(0,3).

Tip

Transformations of y=x2y=x^2: y=a(x−h)2+ky=a(x-h)^2+k shifts the graph right hh, up kk, and stretches by aa (reflects if a<0a<0). Bigger ∣a∣|a| means a narrower parabola.

Solving by Factoring and the Square-Root Method

Concept
Two quick methods

Factoring uses the Zero Product Property: if A⋅B=0A\cdot B=0 then A=0A=0 or B=0B=0. Set the quadratic equal to 00, factor, and solve each factor.

Square-root method works when there is no middle (bxbx) term, or when one side is a perfect square: isolate the square, then take ±\pm the square root of both sides.

Example
Factoring

Solve x2−5x+6=0x^2-5x+6=0. Factor: (x−2)(x−3)=0(x-2)(x-3)=0, so x=2x=2 or x=3x=3.

Example
Square-root method

Solve (x−4)2=25(x-4)^2=25. Take roots: x−4=±5x-4=\pm5, so x=9x=9 or x=−1x=-1.

Tip

Remember the ±\pm! A square root gives two solutions. Dropping the negative root loses half your answers.

Completing the Square

Concept
Building a perfect square
xy

To complete the square on x2+bxx^2+bx, add (b2)2\left(\dfrac{b}{2}\right)^2. This makes

x2+bx+(b2)2=(x+b2)2.x^2+bx+\left(\tfrac{b}{2}\right)^2=\left(x+\tfrac{b}{2}\right)^2.

Use this to solve any quadratic or to convert to vertex form.

The vertex is the turning point.

Example
Completing the square (fully worked)

Solve x2+6x+5=0x^2+6x+5=0.

x2+6x=−5move constant rightx2+6x+9=−5+9add (6/2)2=9 to both sides(x+3)2=4write left as a squarex+3=±2take square rootsx=−3±2  ⇒  x=−1 or x=−5.\begin{aligned} x^2+6x &= -5 &&\text{move constant right}\\ x^2+6x+9 &= -5+9 &&\text{add }(6/2)^2=9\text{ to both sides}\\ (x+3)^2 &= 4 &&\text{write left as a square}\\ x+3 &= \pm2 &&\text{take square roots}\\ x &= -3\pm2 \;\Rightarrow\; x=-1\ \text{or}\ x=-5. \end{aligned}
Example
Converting to vertex form

Write y=x2−4x+7y=x^2-4x+7 in vertex form. Take (4/2)2=4(4/2)^2=4:

y=(x2−4x+4)+7−4=(x−2)2+3,vertex (2,3).y=(x^2-4x+4)+7-4=(x-2)^2+3,\quad\text{vertex }(2,3).
Tip

Tip: If a≠1a\neq1, factor aa out of the x2x^2 and xx terms first, then complete the square inside the parentheses.

The Quadratic Formula and the Discriminant

Concept
The formula that always works

For ax2+bx+c=0ax^2+bx+c=0 with a≠0a\neq0,

x=−b±b2−4ac2a.x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.

The quantity under the radical, D=b2−4acD=b^2-4ac, is the discriminant:

  • D>0D>0: two distinct real solutions.
  • D=0D=0: one repeated real solution.
  • D<0D<0: two complex (nonreal) solutions.
Example
Quadratic formula (fully worked)

Solve 2x2+3x−2=02x^2+3x-2=0, so a=2, b=3, c=−2a=2,\ b=3,\ c=-2.

x=−3±32−4(2)(−2)2(2)=−3±9+164=−3±254=−3±54.\begin{aligned} x&=\frac{-3\pm\sqrt{3^2-4(2)(-2)}}{2(2)} =\frac{-3\pm\sqrt{9+16}}{4} =\frac{-3\pm\sqrt{25}}{4}=\frac{-3\pm5}{4}. \end{aligned}

So x=24=12x=\dfrac{2}{4}=\dfrac12 or x=−84=−2x=\dfrac{-8}{4}=-2. (Here D=25>0D=25>0: two real roots.)

Tip

Check first: compute the discriminant before solving. It tells you how many and what kind of answers to expect.

Complex Numbers: the Imaginary Unit ii

Concept
A new number

The imaginary unit is defined by i=−1i=\sqrt{-1}, so that i2=−1i^2=-1. A complex number has standard form a+bia+bi, where aa is the real part and bb is the imaginary part.

Powers of ii cycle with period 44:

i1=i,i2=−1,i3=−i,i4=1,i5=i, …i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1,\quad i^5=i,\ \dots
Example
Simplifying

−25=25⋅−1=5i\sqrt{-25}=\sqrt{25}\cdot\sqrt{-1}=5i.   i23=i20⋅i3=(i4)5⋅i3=1⋅(−i)=−ii^{23}=i^{20}\cdot i^{3}=(i^4)^5\cdot i^3=1\cdot(-i)=-i.

Tip

Powers of ii: divide the exponent by 44 and keep the remainder: remainder 0→10\to1, 1→i1\to i, 2→−12\to-1, 3→−i3\to-i.

Operations with Complex Numbers

Concept
Add, subtract, multiply, divide

Add/subtract: combine real parts and imaginary parts separately.

Multiply: use FOIL, then replace i2i^2 with −1-1.

Conjugate: the conjugate of a+bia+bi is a−bia-bi. Their product (a+bi)(a−bi)=a2+b2(a+bi)(a-bi)=a^2+b^2 is a real number.

Divide: multiply numerator and denominator by the conjugate of the denominator.

Example
Multiplying

(2+3i)(1−i)=2−2i+3i−3i2=2+i−3(−1)=5+i.(2+3i)(1-i)=2-2i+3i-3i^2=2+i-3(-1)=5+i.

Example
Dividing (using the conjugate)
3+2i1−i=3+2i1−i⋅1+i1+i=3+3i+2i+2i21−i2=1+5i2=12+52i.\frac{3+2i}{1-i}=\frac{3+2i}{1-i}\cdot\frac{1+i}{1+i} =\frac{3+3i+2i+2i^2}{1-i^2} =\frac{1+5i}{2}=\frac12+\frac52 i.
Tip

Tip: A complex answer is not finished until it is in the form a+bia+bi. Real and imaginary parts should be separated.

Quadratics with Complex Solutions

Concept
When the parabola misses the xx-axis
xy

If the discriminant D<0D<0, the graph has no xx-intercepts and the two solutions are complex conjugates p±qip\pm qi. Solve exactly as before---the negative under the radical just introduces ii.

The vertex is the turning point.

Example
Complex roots (fully worked)

Solve x2−2x+5=0x^2-2x+5=0, so a=1, b=−2, c=5a=1,\ b=-2,\ c=5.

x=2±(−2)2−4(1)(5)2=2±4−202=2±−162=2±4i2=1±2i.\begin{aligned} x&=\frac{2\pm\sqrt{(-2)^2-4(1)(5)}}{2} =\frac{2\pm\sqrt{4-20}}{2} =\frac{2\pm\sqrt{-16}}{2}=\frac{2\pm4i}{2}=1\pm2i. \end{aligned}

The solutions are 1+2i1+2i and 1−2i1-2i, a conjugate pair.

Tip

Remember: complex roots always come in conjugate pairs p+qip+qi and p−qip-qi (for real coefficients).

Modeling with Quadratics

Concept
Where parabolas appear
xy

Quadratics model projectile height, areas, and any quantity with a single maximum or minimum.

  • Projectiles: h(t)=−16t2+v0t+h0h(t)=-16t^2+v_0t+h_0 (feet, seconds). The peak is at the vertex time t=−v02(−16)t=-\dfrac{v_0}{2(-16)}; the ground is where h(t)=0h(t)=0.
  • Max/min: the extreme value always occurs at the vertex.

The vertex is the turning point.

Example
Projectile

A ball is thrown so h(t)=−16t2+32th(t)=-16t^2+32t. Vertex time: t=−322(−16)=1t=-\dfrac{32}{2(-16)}=1 s. Max height h(1)=−16+32=16h(1)=-16+32=16 ft. It lands when −16t2+32t=0⇒−16t(t−2)=0-16t^2+32t=0\Rightarrow-16t(t-2)=0, at t=2t=2 s.

Tip

Tip: “When is it highest?” asks for the vertex time; “how high?” asks for the vertex yy-value; “when does it land?” asks for the positive root.

Going Deeper: Advanced Quadratics & Complex Numbers

Concept
Vieta's formulas: roots without solving
xy

For ax2+bx+c=0ax^2+bx+c=0 with roots r1,r2r_1,r_2, dividing by aa gives x2+bax+ca=0x^2+\dfrac{b}{a}x+\dfrac{c}{a}=0. Comparing with (x−r1)(x−r2)=x2−(r1+r2)x+r1r2(x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1r_2 shows

r1+r2=−ba,r1r2=ca.r_1+r_2=-\frac{b}{a},\qquad r_1 r_2=\frac{c}{a}.

These Vieta's formulas let you read the sum and product of the roots straight off the coefficients---no solving required. Any symmetric function of the roots can then be rebuilt from these two numbers, e.g.

r12+r22=(r1+r2)2−2r1r2,1r1+1r2=r1+r2r1r2.r_1^2+r_2^2=(r_1+r_2)^2-2r_1r_2,\qquad \frac{1}{r_1}+\frac{1}{r_2}=\frac{r_1+r_2}{r_1 r_2}.

The vertex is the turning point.

Example
Symmetric functions of roots (fully worked)

Without solving, evaluate r12+r22r_1^2+r_2^2 and 1r1+1r2\dfrac{1}{r_1}+\dfrac{1}{r_2} for 2x2−5x+1=02x^2-5x+1=0.

First read off the symmetric quantities with a=2, b=−5, c=1a=2,\ b=-5,\ c=1:

r1+r2=−ba=52,r1r2=ca=12.r_1+r_2=-\frac{b}{a}=\frac{5}{2},\qquad r_1 r_2=\frac{c}{a}=\frac{1}{2}.

Now assemble the answers from these:

r12+r22=(r1+r2)2−2r1r2=(52)2−2⋅12=254−1=214,1r1+1r2=r1+r2r1r2=5/21/2=5.\begin{aligned} r_1^2+r_2^2 &= (r_1+r_2)^2-2r_1r_2 = \left(\tfrac52\right)^2-2\cdot\tfrac12 = \tfrac{25}{4}-1 = \tfrac{21}{4},\\[2pt] \frac{1}{r_1}+\frac{1}{r_2} &= \frac{r_1+r_2}{r_1 r_2}=\frac{5/2}{1/2}=5. \end{aligned}
Tip

Test for a symmetric expression: if swapping r1↔r2r_1\leftrightarrow r_2 leaves it unchanged, it can be written using only r1+r2r_1+r_2 and r1r2r_1r_2. Start every such problem by writing down those two values.

Concept
Building a quadratic from transformed roots

To construct a monic quadratic with prescribed roots s1,s2s_1,s_2, use

x2−(s1+s2) x+s1s2=0.x^2-(s_1+s_2)\,x+s_1 s_2=0.

When the new roots are transformations of old roots r1,r2r_1,r_2, compute the new sum and product from Vieta's formulas of the original equation---you never need the roots themselves.

  • Roots r1+k, r2+kr_1+k,\ r_2+k: replace xx by x−kx-k in the original.
  • Roots kr1, kr2kr_1,\ kr_2: new sum k(r1+r2)k(r_1+r_2), new product k2r1r2k^2 r_1r_2.
  • Roots 1r1,1r2\dfrac{1}{r_1},\dfrac{1}{r_2}: reverse the coefficient list a,b,c→c,b,aa,b,c\to c,b,a.
Example
New quadratic whose roots are the squares (fully worked)

The equation x2−6x+4=0x^2-6x+4=0 has roots r1,r2r_1,r_2. Build a quadratic whose roots are r12r_1^2 and r22r_2^2.

From Vieta: r1+r2=6r_1+r_2=6 and r1r2=4r_1r_2=4. The new sum and product are

r12+r22=(r1+r2)2−2r1r2=36−8=28,r12 r22=(r1r2)2=16.\begin{aligned} r_1^2+r_2^2 &= (r_1+r_2)^2-2r_1r_2 = 36-8 = 28,\\ r_1^2\,r_2^2 &= (r_1r_2)^2 = 16. \end{aligned}

So a quadratic with these roots is

x2−28x+16=0.x^2-28x+16=0.
Concept
Discriminant & parameter conditions

Treat the discriminant D=b2−4acD=b^2-4ac as a function of an unknown parameter to force a desired root behavior:

  • Two distinct real roots: D>0D>0.
  • One repeated (double) root: D=0D=0.
  • No real roots (complex pair): D<0D<0.

Setting D=0D=0 also gives the tangency condition when a line meets a parabola in exactly one point (see below), and the boundary value where the number of real roots changes.

Tip

Watch the leading coefficient. If the parameter multiplies x2x^2, first insist a≠0a\neq0 (otherwise the equation is linear, not quadratic) before applying a discriminant condition.

Example
Line tangent to a parabola (fully worked)

For which value of mm is the line y=mx+2y=mx+2 tangent to the parabola y=x2+3x+6y=x^2+3x+6?

Tangency means the system meets in exactly one point, so set the two expressions equal and require a double root:

x2+3x+6=mx+2x2+(3−m)x+4=0.\begin{aligned} x^2+3x+6 &= mx+2\\ x^2+(3-m)x+4 &= 0. \end{aligned}

A double root needs D=0D=0:

D=(3−m)2−4(1)(4)=0(3−m)2=163−m=±4  ⇒  m=−1 or m=7.\begin{aligned} D=(3-m)^2-4(1)(4) &= 0\\ (3-m)^2 &= 16\\ 3-m &= \pm4 \;\Rightarrow\; m=-1 \ \text{or}\ m=7. \end{aligned}

Both lines touch the parabola at a single point.

Concept
Modulus & argument of a complex number

Plot z=a+biz=a+bi as the point (a,b)(a,b). Its modulus (distance from the origin) and argument (angle from the positive real axis) are

∣z∣=a2+b2,θ=arg⁡z=atan2⁡(b,a).|z|=\sqrt{a^2+b^2},\qquad \theta=\arg z=\operatorname{atan2}(b,a).

Key facts: ∣z∣2=zzˉ|z|^2=z\bar z, moduli multiply (∣z1z2∣=∣z1∣∣z2∣|z_1 z_2|=|z_1||z_2|), and multiplying by ii rotates a point 90∘90^\circ counterclockwise. This geometric view explains why the powers of ii cycle: ii has modulus 11 and argument 90∘90^\circ, so ini^n walks around the unit circle in 90∘90^\circ steps.

Example
Modulus and argument (fully worked)

Find the modulus and argument of z=−1+i3z=-1+i\sqrt3.

∣z∣=(−1)2+(3)2=1+3=2,θ=atan2⁡(3,−1)=120∘,\begin{aligned} |z| &= \sqrt{(-1)^2+(\sqrt3)^2}=\sqrt{1+3}=2,\\ \theta &= \operatorname{atan2}(\sqrt3,-1)=120^\circ, \end{aligned}

because (−1,3)(-1,\sqrt3) sits in the second quadrant. So zz has modulus 22 and argument 120∘120^\circ.

Example
Solving z2=zˉz^2=\bar z (fully worked)

Find every complex z=a+biz=a+bi with z2=zˉz^2=\bar z.

Take moduli of both sides: ∣z∣2=∣zˉ∣=∣z∣|z|^2=|\bar z|=|z|, so ∣z∣(∣z∣−1)=0|z|(|z|-1)=0. Hence ∣z∣=0|z|=0 or ∣z∣=1|z|=1.

  • If ∣z∣=0|z|=0 then z=0z=0, which checks: 02=00^2=0.
  • If ∣z∣=1|z|=1, multiply z2=zˉz^2=\bar z by zz: z3=zzˉ=∣z∣2=1z^3=z\bar z=|z|^2=1. So zz is a cube root of unity.

The cube roots of unity are z=1z=1 and z=−12±32iz=-\tfrac12\pm\tfrac{\sqrt3}{2}i. Together with z=0z=0 these are the four solutions:

z=0,z=1,z=−12+32i,z=−12−32i.z=0,\quad z=1,\quad z=-\tfrac12+\tfrac{\sqrt3}{2}i,\quad z=-\tfrac12-\tfrac{\sqrt3}{2}i.
Tip

Conjugate root pairs. A polynomial with real coefficients has complex roots only in conjugate pairs: if p+qip+qi is a root, so is p−qip-qi. That pair multiplies back to the real quadratic factor

(x−(p+qi))(x−(p−qi))=x2−2px+(p2+q2),\bigl(x-(p+qi)\bigr)\bigl(x-(p-qi)\bigr)=x^2-2px+(p^2+q^2),

whose sum of roots 2p2p and product p2+q2=∣p+qi∣2p^2+q^2=|p+qi|^2 are exactly Vieta's formulas again.

Reminder — Vieta's formulas:  r1+r2=−ba,r1r2=car_1+r_2=-\frac{b}{a},\qquad r_1 r_2=\frac{c}{a}

Formulas, Proofs & Tips (1 of 2)

Tip
The quadratic formula
x=−b±b2−4ac2a(a≠0)x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad (a\neq 0)

What it means. Solves every quadratic ax2+bx+c=0ax^2+bx+c=0, factorable or not.

Example. x2−5x+6=0x^2-5x+6=0: x=5±25−242x=\dfrac{5\pm\sqrt{25-24}}{2}, so x=3x=3 or x=2x=2.

Why it works. Complete the square. Divide by aa: x2+bax=−cax^2+\tfrac{b}{a}x=-\tfrac{c}{a}. Add (b2a)2\left(\tfrac{b}{2a}\right)^2 to both sides so the left is a perfect square: (x+b2a)2=b2−4ac4a2\left(x+\tfrac{b}{2a}\right)^2=\tfrac{b^2-4ac}{4a^2}. Take square roots and subtract b2a\tfrac{b}{2a}.

Tip. Write the equation as ax2+bx+c=0ax^2+bx+c=0 first — every term must be on one side, or your a,b,ca,b,c will be wrong.

Tip
The discriminant
Δ=b2−4ac\Delta = b^{2}-4ac

What it means. The part under the root. Δ>0\Delta>0: two real roots. Δ=0\Delta=0: one repeated root. Δ<0\Delta<0: two complex roots.

Example. x2−4x+4x^2-4x+4 has Δ=16−16=0\Delta=16-16=0, so one repeated real root.

Why it works. In the quadratic formula the roots differ by ±Δ2a\tfrac{\pm\sqrt{\Delta}}{2a}. A positive Δ\Delta gives two distinct real shifts, zero gives none, and a negative Δ\Delta forces the square root to be imaginary.

Tip. If Δ\Delta is a perfect square the quadratic factors over the integers — worth checking before reaching for the formula.

Tip
Vertex of a parabola
x=−b2a,y=f ⁣(−b2a)x=-\frac{b}{2a}, \qquad y=f\!\left(-\frac{b}{2a}\right)

What it means. The turning point — the minimum if a>0a>0, the maximum if a<0a<0.

Example. y=x2−6x+5y=x^2-6x+5: vertex at x=−−62=3x=-\tfrac{-6}{2}=3, y=−4y=-4, so (3,−4)(3,-4).

Why it works. The two roots of the quadratic formula sit symmetrically either side of −b2a-\tfrac{b}{2a}, so that value is the axis of symmetry. A parabola turns exactly on its axis of symmetry.

Tip. This also gives the max/min value of a quadratic without calculus — useful for word problems about greatest area or lowest cost.

Tip
Complex number arithmetic
i2=−1,(a+bi)(c+di)=(ac−bd)+(ad+bc)ii^2=-1,\qquad (a+bi)(c+di)=(ac-bd)+(ad+bc)i

What it means. ii is the number whose square is −1-1; complex numbers multiply like binomials, using i2=−1i^2=-1 at the end.

Example. (2+i)(3−i)=6−2i+3i−i2=7+i(2+i)(3-i)=6-2i+3i-i^2=7+i.

Why it works. Expand (a+bi)(c+di)=ac+adi+bci+bd i2(a+bi)(c+di)=ac+adi+bci+bd\,i^2 and replace i2i^2 by −1-1, which turns bd i2bd\,i^2 into −bd-bd and collects into real and imaginary parts.

Tip. To divide, multiply top and bottom by the conjugate c−dic-di: (c+di)(c−di)=c2+d2(c+di)(c-di)=c^2+d^2 is real, clearing ii from the denominator.

Formulas, Proofs & Tips (2 of 2)

Tip
Vieta's formulas
r1+r2=−ba,r1r2=car_1+r_2=-\frac{b}{a},\qquad r_1 r_2=\frac{c}{a}

What it means. The sum and product of a quadratic's roots read straight off its coefficients.

Example. x2−7x+12=0x^2-7x+12=0: the roots sum to 77 and multiply to 1212, so they are 33 and 44.

Why it works. If r1,r2r_1,r_2 are the roots then ax2+bx+c=a(x−r1)(x−r2)=a(x2−(r1+r2)x+r1r2)ax^2+bx+c=a(x-r_1)(x-r_2)=a\left(x^2-(r_1+r_2)x+r_1r_2\right). Matching coefficients gives both formulas.

Tip. Great for "find the sum of the roots" questions — you never have to solve the equation.