Math 2–3 Honors · Lesson Review · October 15, 2026

Factored Form and Quadratic Graphs

Read zeros from factors, find the vertex from symmetry, connect three quadratic forms, and handle irrational zeros or no real x-intercepts.

29-minute edited lesson9 chaptersCaptions + transcript

01 · Watch

Lesson video

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0:0001Graph a quadratic from factored form 4:2302Read the zeros from the factors 6:0903The midpoint gives the axis 8:1704Factored form and the meaning of a zero 10:3205A vertical stretch: find the vertex 16:0106Find an equation from a graph 18:2107Find irrational zeros 21:4108Turn exact zeros into factors 25:3309When there are no real zeros

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02 · Understand

Learning targets

Read a quadratic’s real zeros from factored form.
Use the midpoint of the zeros to find the axis of symmetry and then the vertex.
Connect factored, standard, and vertex forms without changing the function.
Find a factored equation from a graph, including its vertical scale, and use the quadratic formula for irrational zeros.

03 · Connect

Key ideas

This lesson connects zeros, symmetry, the vertex, and three useful forms of the same quadratic. All graphs and factorizations below are over the real numbers unless stated otherwise.

01A zero is an input, not an ordered pair: $f(r)=0\quad\Longleftrightarrow\quad(r,0)\text{ is an }x\text{-intercept}$ A zero makes the function value equal zero. In f(x)=(x−1)(x+3), the zeros are 1 and −3; the intercept points are (1,0) and (−3,0).
02Use the opposite sign inside each factor: $x-r=0\quad\Longrightarrow\quad x=r$ Set each factor equal to zero. The factor x+3 corresponds to the zero −3. The constant coefficient a is nonzero and does not add another zero.
03The axis lies halfway between the zeros: $h=\frac{r_1+r_2}{2},\qquad\text{axis: }x=h,\qquad\text{vertex: }(h,f(h))$ Average the x-coordinates, not their absolute values. This agrees with h=−b/(2a) after expansion. The midpoint formula applies when the zeros are real; they may be equal.
04Keep the leading coefficient: $f(h+d)=k+ad^2$ Starting at the vertex, use horizontal offsets ±1, ±2, and ±3 with vertical offsets a, 4a, and 9a. Positive a opens upward; negative a opens downward. The offsets are measured from the vertex.
05The forms emphasize different features: $\underbrace{a(x-r_1)(x-r_2)}_{\text{zeros}}=\underbrace{ax^2+bx+c}_{\text{standard form}}=\underbrace{a(x-h)^2+k}_{\text{vertex}}$ Use a form that displays the feature you need. Expand to check a factorization, substitute x=0 for the y-intercept, and keep the same leading coefficient through every conversion.
06Irrational zeros still produce exact real factors: $r_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ If an integer factor pair is unavailable, solve f(x)=0 with the quadratic formula. A positive discriminant gives two real zeros; a zero discriminant gives one repeated real zero. A negative discriminant gives no real x-intercepts and no real linear factorization.
07A third point determines the scale: $y_0=a(x_0-r_1)(x_0-r_2)$ Two zeros alone do not determine a unique parabola. Substitute another point that is not a zero to solve for a. For any quadratic, the domain is all real numbers; its range begins or ends at the vertex height.

04 · Apply

Worked examples

These examples follow the methods developed in the recorded lesson.

Classroom example

From factored form to the first graph

f(x)=(x-1)(x+3)=x^2+2x-3

The zeros are 1 and −3. Expanding provides the standard-form coefficients.

h=-\frac{2}{2(1)}=-1,\qquad f(-1)=(-2)(2)=-4

The standard-form vertex formula gives the vertex (−1,−4). The midpoint of the zeros gives the same h.

f(x)=(x+1)^2-4

Graph the vertex, the symmetric zeros (−3,0) and (1,0), and matching nearby points. The parabola opens upward and its range is y≥−4.

Classroom example

Use two zeros and a vertical stretch

f(x)=2(x-2)(x+4),\quad r_1=2,\quad r_2=-4

The factor 2 changes vertical scale but leaves the zeros at 2 and −4.

h=\frac{2+(-4)}{2}=-1,\qquad k=f(-1)=2(-3)(3)=-18

Average the zeros, then evaluate the original function at their midpoint.

f(x)=2(x+1)^2-18=2x^2+4x-16

The vertex is (−1,−18). From that vertex, the first three vertical offsets are 2, 8, and 18.

Classroom example

Recover a downward-opening equation from a graph

r_1=-3,\quad r_2=-1,\qquad f(x)=a(x+3)(x+1)

The zeros are −3 and −1. The graph’s vertex is (−2,2), and a one-unit horizontal step from it reaches (−1,0).

0=2+a(1)^2\quad\Longrightarrow\quad a=-2

From the vertex, one step right moves down two units. This gives a=−2, just as substituting the y-intercept (0,−6) into factored form would.

h=\frac{-3+(-1)}{2}=-2,\qquad f(-2)=-2(1)(-1)=2

The vertex is (−2,2); the axis is x=−2.

f(x)=-2(x+1)(x+3)=-2(x+2)^2+2=-2x^2-8x-6

The negative coefficient makes the graph open downward. All three forms describe the same function.

Classroom example

Factor a quadratic with irrational zeros

f(x)=x^2+2x-1,\qquad x^2+2x-1=0

No integer pair multiplies to −1 and adds to 2. Solve for the zeros instead.

x=\frac{-2\pm\sqrt{2^2-4(1)(-1)}}{2}=\frac{-2\pm\sqrt8}{2}=-1\pm\sqrt2

Simplify the radical and divide both numerator terms by 2.

f(x)=\bigl(x+1-\sqrt2\bigr)\bigl(x+1+\sqrt2\bigr)

Each factor is x minus one of the zeros. Expanding the conjugate pair gives (x+1)²−2.

f(x)=(x+1)^2-2,\qquad\text{vertex }(-1,-2)

The roots are approximately −2.414 and 0.414. Their midpoint is −1, agreeing with the vertex form.

Classroom example

A quadratic with no real linear factors

f(x)=(x-3)^2+2=x^2-6x+11

The minimum value is 2, so the graph never reaches the x-axis.

b^2-4ac=(-6)^2-4(1)(11)=36-44=-8

The negative discriminant confirms that there are no real zeros.

\text{domain: }\mathbb R,\qquad\text{range: }[2,\infty)

It has standard and vertex forms but no factorization into real linear factors. Complex factorization is a different question.

05 · Avoid

Common mistakes

Copying a factor’s sign as the zero.

Set the factor equal to zero first. For x+4, the zero is −4.

Averaging without the negative signs.

For zeros 2 and −4, calculate (2−4)/2=−1, not (2+4)/2.

Stopping after the midpoint.

The midpoint gives only h. Calculate k=f(h) to finish the vertex.

Dropping the coefficient a.

Zeros do not give the vertical scale. Preserve a in factored form and in the vertex-relative step pattern.

Confusing zeros with intercept points.

Zeros are x-values. An x-intercept is an ordered pair with y=0.

Treating “no integer factors” as “no real zeros.”

Use the discriminant and the quadratic formula. Irrational zeros can give exact real linear factors.

06 · Check yourself

Try it first

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

1Find the zeros, axis, vertex, opening direction, and range:

f(x)=(x-4)(x+2)

$r_1=4,\quad r_2=-2,\quad h=\frac{4-2}{2}=1$

The intercepts are (4,0) and (−2,0).

$k=f(1)=(-3)(3)=-9,\qquad f(x)=(x-1)^2-9$

The vertex is (1,−9), the axis is x=1, and the parabola opens upward. The domain is ℝ and the range is y≥−9.

2Graph using the zeros and vertex:

g(x)=-2(x+3)(x-1)

$r_1=-3,\quad r_2=1,\quad h=\frac{-3+1}{2}=-1$

Set the two factors equal to zero.

$k=g(-1)=-2(2)(-2)=8,\qquad g(x)=-2(x+1)^2+8$

The vertex is (−1,8) and the axis is x=−1. The graph opens downward, passes through (0,6), and has range y≤8.

3Factor, then find the vertex without completing the square:

p(x)=2x^2-4x-6

$p(x)=2(x^2-2x-3)=2(x-3)(x+1)$

The zeros are 3 and −1.

$h=\frac{3-1}{2}=1,\qquad k=p(1)=2-4-6=-8$

The vertex is (1,−8), the axis is x=1, and vertex form is p(x)=2(x−1)²−8.

4Use exact zeros to write real factored form, then identify the vertex:

q(x)=x^2-4x+1

$x=\frac{4\pm\sqrt{16-4}}{2}=2\pm\sqrt3$

The two real zeros are irrational.

$q(x)=\bigl(x-2-\sqrt3\bigr)\bigl(x-2+\sqrt3\bigr)$

Write x minus each zero, keeping the grouping signs.

$h=2,\quad k=q(2)=-3,\qquad q(x)=(x-2)^2-3$

The vertex is (2,−3). Expanding either expression gives x²−4x+1.

5A parabola has zeros −2 and 6 and passes through (0,−24). Find its factored equation and vertex.

f(x)=a(x+2)(x-6)

$-24=a(2)(-6)=-12a\quad\Longrightarrow\quad a=2$

The extra point determines the vertical scale.

$f(x)=2(x+2)(x-6),\quad h=\frac{-2+6}{2}=2$

Average the zeros for the axis.

$k=f(2)=2(4)(-4)=-32,\qquad f(x)=2(x-2)^2-32$

The vertex is (2,−32). The domain is ℝ and the range is y≥−32.

07 · Revisit

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