Math 2–3 Honors · Lesson Review · October 22, 2026

Complex Zeros and Quadratic Applications

Use conjugate zeros to reconstruct real quadratics, decide when a scale factor is determined, build a quadratic from its vertex and a point, solve intersections, and find a line through two vertices.

29-minute edited lesson12 chaptersCaptions + transcript

01 · Watch

Lesson video

VTT captions SRT captions

0:0001Why nonreal zeros come in pairs 5:2702Reconstruct a quadratic from 2i 9:4103Conjugate factors and an unknown scale 12:4504Is the axis enough information?13:4605Add a leading coefficient 14:3306Use a vertex and a point 16:4707Expand the complete quadratic 17:3108Intersections are systems 20:0309Intersect two quadratics 22:2610Solve for x, then find ordered pairs 25:0711A line through two vertices 27:1512Slope and point-slope form

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02 · Understand

Learning targets

Use a negative discriminant to identify a pair of nonreal conjugate zeros of a real quadratic.
Build a quadratic from its zeros and a scale condition, and recognize when the information leaves a family of functions.
Use a vertex and one additional point to write a quadratic in vertex and standard forms.
Find intersections by setting function values equal, then report matching ordered pairs.
Find vertices and use them to write the equation of a line.

03 · Connect

Key ideas

The lesson moves from complex zeros to complete quadratic models, then uses familiar forms to solve intersections and connect vertices. Throughout, coefficients of the quadratics are real and the leading coefficient is nonzero.

01Read the discriminant: $D=b^2-4ac,\qquad x=\frac{-b\pm\sqrt D}{2a}$ For real a, b, and c with a≠0, a negative discriminant gives two nonreal conjugate solutions. Zero gives a repeated real solution; a positive discriminant gives two distinct real solutions.
02Conjugates share a real part: $z=u+vi,\qquad\overline z=u-vi\quad(u,v\in\mathbb R)$ For a polynomial with real coefficients, nonreal zeros occur in conjugate pairs. The points are mirror images across the real axis in the complex plane. This conclusion needs real coefficients.
03Turn zeros into factors: $f(x)=a(x-(u+vi))(x-(u-vi))=a\bigl((x-u)^2+v^2\bigr)$ Subtract each entire zero from x. Group the imaginary terms as conjugates before multiplying. Use i²=−1 so the product has real coefficients.
04A zero pair does not determine the scale: $f(x)=a\bigl((x-u)^2+v^2\bigr),\quad a\ne0,\quad\text{axis }x=u$ The axis is already fixed by the real part of the conjugate zeros. Repeating that axis as extra information does not determine a. A leading coefficient or a suitable point is needed for one specific quadratic.
05Use the vertex efficiently: $f(x)=a(x-h)^2+k,\qquad a=\frac{y_0-k}{(x_0-h)^2}\quad(x_0\ne h)$ The vertex gives h and k immediately. Substitute a different input-output point to find a, then expand if standard form is requested. A valid quadratic requires a≠0.
06Intersections require the same input and output: $f(x)=g(x),\qquad (x,y)=(x,f(x))=(x,g(x))$ Solving f(x)=g(x) finds candidate x-coordinates. Evaluate either function at each input and check the same input in the other function. Do not compare outputs computed at different x-values.
07A line through two vertices is an ordinary two-point line: $m=\frac{y_2-y_1}{x_2-x_1},\qquad y-y_1=m(x-x_1)\quad(x_2\ne x_1)$ First find the two vertices. Then use slope and point-slope form. If the vertices have the same x-coordinate, their connecting line is vertical and has equation x=x₁.

04 · Apply

Worked examples

These examples follow the methods developed in the recorded lesson.

Classroom example · conjugate zeros

A zero of 2i and leading coefficient −3

2i\text{ is a zero}\quad\Longrightarrow\quad -2i\text{ is a zero}

The quadratic has real coefficients, so use the conjugate partner.

f(x)=-3(x-2i)(x+2i)

Keep the given leading coefficient as the multiplier.

(x-2i)(x+2i)=x^2-(2i)^2=x^2+4

The middle terms cancel and i²=−1.

\boxed{f(x)=-3x^2-12}

Substituting either 2i or −2i gives zero.

Classroom example · enough information?

A zero of 3+2i and axis x=3

3+2i,\quad3-2i

Real coefficients supply the second zero.

f(x)=a(x-(3+2i))(x-(3-2i))

A pair of zeros determines factors, but the scale a is still unknown.

f(x)=a\bigl((x-3)^2+4\bigr)=ax^2-6ax+13a,\quad a\ne0

This is the full family of real quadratics satisfying the given conditions.

-\frac{b}{2a}=-\frac{-6a}{2a}=3

Every member already has axis x=3. That condition does not give a new equation for a. One unique standard-form polynomial cannot be determined without another scale condition.

a=3\quad\Longrightarrow\quad f(x)=3x^2-18x+39

Later in class, the teacher adds the leading coefficient 3 as a new assumption. That selects this one member; the original zero-and-axis data alone do not.

Classroom example · vertex and point

Vertex (−3,2), passing through (−1,14)

f(x)=a(x+3)^2+2

Use vertex form with h=−3 and k=2.

14=a(-1+3)^2+2=4a+2\quad\Longrightarrow\quad a=3

Substitute the given point.

f(x)=3(x+3)^2+2=\boxed{3x^2+18x+29}

Checked correction to the expanded board answer: distribute 3 to every term, including 6x. The correct middle term is 18x. This model gives f(−3)=2 and f(−1)=14.

Classroom setup · two lines

Why setting two expressions equal finds an intersection

y=2x+3,\qquad y=-3x-7

At an intersection, both expressions represent the same y-value.

2x+3=-3x-7\quad\Longrightarrow\quad x=-2

Checked continuation of the classroom setup: keep the original constant −7 when setting the expressions equal, then collect like terms.

y=2(-2)+3=-1=-3(-2)-7

Both functions agree at the same input.

\boxed{(-2,-1)}

This algebraic method extends to nonlinear graphs.

Classroom example · two parabolas

Intersect x²+2x−2 and −x²+4x+10

x^2+2x-2=-x^2+4x+10

Set the function values equal.

2x^2-2x-12=0\quad\Longleftrightarrow\quad(x-3)(x+2)=0

Divide by 2, then factor.

x=3\text{ or }x=-2

The equation gives two candidate inputs.

f(3)=13=g(3),\qquad f(-2)=-2=g(-2)

Check corresponding outputs, using the same input in both functions.

\boxed{(3,13)\text{ and }(-2,-2)}

Report full ordered pairs, rather than only the two x-values.

Classroom example · a line joining vertices

Connect the vertices of a quadratic and an absolute-value graph

f(x)=x^2+4x+1=(x+2)^2-3,\qquad g(x)=2|x-3|+1

The vertices are (−2,−3) and (3,1).

m=\frac{1-(-3)}{3-(-2)}=\frac45

Use change in y divided by change in x. The spoken rule is correct; this corrects the handwritten denominator’s last subscripted letter.

y-1=\frac45(x-3)

Point-slope form uses the vertex (3,1).

\boxed{y=\frac45x-\frac75}

Both vertices satisfy this line. It joins the vertices; it is not a statement that the original graphs intersect at those vertices.

05 · Avoid

Common mistakes

Forgetting the real-coefficient condition.

Conjugate pairing follows from real polynomial coefficients. A complex-coefficient polynomial need not contain the conjugate partner.

Writing the wrong factor sign.

A zero of 3+2i gives x−(3+2i), which is x−3−2i. Parentheses prevent a lost minus sign.

Using i²=1.

The imaginary unit satisfies i²=−1. That sign makes (x−u−vi)(x−u+vi) equal (x−u)²+v².

Assuming a=1 without a reason.

Two zeros determine a quadratic only up to a nonzero multiplier. An axis already implied by those zeros supplies no additional scale information.

Losing the middle term when squaring.

For example, (x+3)²=x²+6x+9. Expand completely before multiplying by the leading coefficient.

Checking different inputs against each other.

Compare f(3) with g(3), and f(−2) with g(−2). Outputs at different inputs do not test an intersection.

Stopping at intersection x-values.

Find the corresponding y-coordinate for each x. The answer to a graph-intersection question is a set of ordered pairs.

06 · Check yourself

Try it first

Use these five fresh problems to check your understanding. Open a solution after you try it.

1Solve the quadratic equation and identify the conjugate pair.

2x^2-4x+10=0

$D=(-4)^2-4(2)(10)=-64$

The discriminant is negative.

$x=\frac{4\pm\sqrt{-64}}4=\frac{4\pm8i}4$

Write the negative square root using i.

$\boxed{x=1\pm2i}$

The solutions share real part 1 and have opposite imaginary parts.

2Write a real quadratic with zero −1+3i and leading coefficient 2.

z=-1+3i,\qquad a=2

$\overline z=-1-3i$

Use the conjugate partner.

$f(x)=2(x+1-3i)(x+1+3i)=2\bigl((x+1)^2+9\bigr)$

Multiply conjugate factors.

$\boxed{f(x)=2x^2+4x+20}$

The coefficient of x² is 2, and both given conjugate zeros satisfy the polynomial.

3Is the zero and axis enough for one quadratic? Give the family, then choose the member passing through (0,34).

4+i\text{ is a zero},\qquad\text{axis }x=4

$f(x)=a\bigl((x-4)^2+1\bigr),\quad a\ne0$

The axis repeats information already contained in the conjugate pair 4±i. The scale is undetermined.

$34=a(16+1)=17a\quad\Longrightarrow\quad a=2$

The added point supplies a scale condition.

$\boxed{f(x)=2(x-4)^2+2=2x^2-16x+34}$

Check f(0)=34 and the conjugate roots.

4Use the vertex and point to write both vertex and standard forms.

\text{vertex }(2,-3),\qquad(0,5)\text{ lies on the graph}

$f(x)=a(x-2)^2-3$

Insert the vertex.

$5=a(0-2)^2-3\quad\Longrightarrow\quad a=2$

Use the additional point.

$\boxed{f(x)=2(x-2)^2-3=2x^2-8x+5}$

The model has vertex (2,−3) and passes through (0,5).

5Find all real intersection points and check both coordinates.

f(x)=x^2,\qquad g(x)=-x^2+4x+6

$x^2=-x^2+4x+6\quad\Longleftrightarrow\quad(x-3)(x+1)=0$

Set the outputs equal, collect terms, divide by 2, and factor.

$x=3\text{ or }x=-1$

These are the intersection inputs.

$f(3)=9=g(3),\qquad f(-1)=1=g(-1)$

Check each input in both original functions.

$\boxed{(3,9)\text{ and }(-1,1)}$

Two real intersections are present.

07 · Revisit

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