Math 2–3 Honors · Lesson Review · October 19, 2026

Quadratic Properties in Three Forms

Connect a quadratic’s graph, standard form, factored form, and vertex form to its domain, range, zeros, intercepts, and increasing or decreasing intervals.

16-minute edited lesson11 chaptersCaptions + transcript

01 · Watch

Lesson video

VTT captions SRT captions

0:0001Domain, range, and the vertex 2:0002Recover vertex form from a graph 3:0403Expand into standard form 4:1804Find exact irrational zeros 6:2005Write real factored form 7:2106Increasing and decreasing intervals 9:2907Why the vertex endpoint can be included 11:0608Start from factored form 12:5509Vertex form and the maximum 13:4110Standard form and the y-intercept 14:5111Board review: no real zeros (no narration)

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02 · Understand

Learning targets

Find a quadratic’s domain, range, intercepts, vertex, and axis of symmetry.
Read useful information from standard, factored, and vertex forms.
Find exact irrational zeros and distinguish them from no real zeros.
State increasing and decreasing intervals using the vertex’s x-coordinate.

03 · Connect

Key ideas

All quadratics here have real coefficients and nonzero leading coefficient. Real factored form exists only when the quadratic has real zeros; standard and vertex forms always exist.

01Start with the domain and the opening: $\operatorname{dom}(f)=\mathbb R,\qquad a>0:\ \text{opens up},\quad a<0:\ \text{opens down}$ A quadratic polynomial accepts every real input. The sign of a determines whether the vertex is a minimum or maximum.
02Read the vertex from vertex form: $f(x)=a(x-h)^2+k,\qquad\text{vertex }(h,k),\quad\text{axis }x=h$ The sign inside the parentheses is opposite the value of h. For example, x+2 means h=−2. The extreme function value is k, while the vertex is the ordered pair (h,k).
03Range uses output values: $a>0:\ [k,\infty),\qquad a<0:\ (-\infty,k]$ The vertex height is included because the function reaches it. Infinity is never an included endpoint.
04Get the vertex from another form: $h=-\frac{b}{2a},\quad k=f(h);\qquad h=\frac{r_1+r_2}{2}\ \text{when real zeros exist}$ Standard form gives h from its coefficients. Factored form with real zeros gives h from their midpoint. Both methods still require evaluating the function to get k.
05Zeros and the y-intercept answer different questions: $f(x)=0:\ x\text{-intercepts};\qquad f(0)=c:\ (0,c)\text{ is the }y\text{-intercept}$ A zero is an input value. An intercept point includes both coordinates. To find a y-intercept in any form, substitute x=0.
06Irrational zeros are still real zeros: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ Keep exact radicals when finding zeros. A positive discriminant gives two real zeros, zero gives a repeated real zero, and a negative discriminant gives no real zeros or real linear factors.
07Increasing and decreasing depend on h, not k: $\begin{array}{c|cc}&(-\infty,h]&[h,\infty)\\\hline a>0&\text{decreasing}&\text{increasing}\\a<0&\text{increasing}&\text{decreasing}\end{array}$ Read the graph from left to right. Following the lesson, include the vertex endpoint in each interval. Increasing and decreasing compare distinct inputs; they are not claims about change at a single point. Some textbooks use open intervals at the vertex.

04 · Apply

Worked examples

These examples follow the methods developed in the recorded lesson.

Classroom example

Read a graph and connect all three forms

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

The graph has vertex (−2,−1), opens upward, and has vertical coefficient 2.

f(x)=2(x^2+4x+4)-1=2x^2+8x+7

Expand the squared binomial and combine constants. The y-intercept is (0,7).

0=2(x+2)^2-1\ \Longrightarrow\ (x+2)^2=\frac12\ \Longrightarrow\ x=-2\pm\frac{\sqrt2}{2}

Both exact zeros are real, even though they are irrational.

f(x)=2\left(x+2-\frac{\sqrt2}{2}\right)\left(x+2+\frac{\sqrt2}{2}\right)

Subtract each zero from x and preserve the leading coefficient 2.

\text{domain: }\mathbb R,\quad\text{range: }[-1,\infty),\quad\text{axis: }x=-2

The minimum value is −1. The graph decreases on (−∞,−2] and increases on [−2,∞).

Classroom example

A downward parabola given in factored form

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

The zeros are −1 and 3, and the negative leading coefficient makes the graph open downward.

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

The vertex is (1,4), with maximum value 4 and axis x=1.

f(x)=-x^2+2x+3=-(x-1)^2+4

All three forms have the same graph. The y-intercept is (0,3).

\text{domain: }\mathbb R,\qquad\text{range: }(-\infty,4]

It increases on (−∞,1] and decreases on [1,∞). Use the vertex’s x-coordinate 1 as the split.

Classroom board example · silent review

Standard form with no real zeros

f(x)=x^2-4x+6,\qquad h=-\frac{-4}{2(1)}=2,\qquad f(2)=2

The vertex is (2,2), the minimum value is 2, and the y-intercept is (0,6).

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

This is vertex form. A squared-binomial expression plus a constant is not a product of two real linear factors.

0=(x-2)^2+2\ \Longrightarrow\ (x-2)^2=-2

No real square equals a negative number. Equivalently, the discriminant is 16−24=−8, so there are no real zeros.

\text{domain: }\mathbb R,\quad\text{range: }[2,\infty),\quad\text{axis: }x=2

The correct interval split is x=2: decreasing on (−∞,2], increasing on [2,∞). This also corrects any temporary board label during the discussion.

05 · Avoid

Common mistakes

Using the wrong sign for h.

In (x+2)², h=−2. In (x−2)², h=2.

Mixing up the vertex and extreme value.

The vertex is (h,k); the minimum or maximum value is only k.

Treating vertex form as factored form.

a(x−h)²+k is vertex form. Real linear factored form is a product a(x−r₁)(x−r₂), and it requires real zeros.

Using k for the interval boundary.

Increasing and decreasing intervals contain x-values. Their boundary is h, the vertex’s horizontal coordinate.

Assuming every quadratic has two real zeros.

Check the graph, completed square, or discriminant. A parabola entirely above the x-axis has no real zeros.

Dropping a during expansion or factorization.

Preserve the leading coefficient. Expanding each form is a useful check that the function stayed the same.

06 · Check yourself

Try it first

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

1Find standard form, exact zeros, vertex, y-intercept, range, and monotonic intervals:

g(x)=3(x-2)^2-6

$g(x)=3x^2-12x+6,\qquad\text{vertex }(2,-6),\quad g(0)=6$

The axis is x=2, the minimum is −6, and the y-intercept is (0,6).

$0=3(x-2)^2-6\ \Longrightarrow\ x=2\pm\sqrt2$

The zeros are 2−√2 and 2+√2. Real factored form is 3(x−2−√2)(x−2+√2).

$\text{range: }[-6,\infty),\quad\text{decreasing: }(-\infty,2],\quad\text{increasing: }[2,\infty)$

The domain is all real numbers.

2Connect the three forms and list the graph’s main properties:

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

$r_1=-3,\quad r_2=1,\quad h=-1,\quad p(-1)=8$

The vertex is (−1,8), with maximum 8 and axis x=−1.

$p(x)=-2x^2-4x+6=-2(x+1)^2+8$

The y-intercept is (0,6), the domain is ℝ, and the range is (−∞,8].

$\text{increasing: }(-\infty,-1],\qquad\text{decreasing: }[-1,\infty)$

The negative coefficient reverses the increasing/decreasing pattern of an upward parabola.

3Write vertex form and determine whether real linear factored form exists:

q(x)=x^2+6x+11

$q(x)=(x+3)^2+2,\quad\text{vertex }(-3,2),\quad\text{axis }x=-3$

The minimum is 2, so the range is [2,∞); the y-intercept is (0,11).

$\Delta=6^2-4(1)(11)=-8<0$

There are no real zeros and no real linear factorization. The domain is ℝ.

$\text{decreasing: }(-\infty,-3],\qquad\text{increasing: }[-3,\infty)$

The interval boundary is −3, not the minimum value 2.

4A parabola has zeros −4 and 2 and y-intercept (0,16). Find its three forms and vertex.

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

$16=a(4)(-2)\ \Longrightarrow\ a=-2$

The additional point fixes the vertical coefficient.

$f(x)=-2(x+4)(x-2)=-2x^2-4x+16$

The axis is halfway between the zeros: x=−1.

$f(-1)=18,\qquad f(x)=-2(x+1)^2+18$

The vertex is (−1,18), its maximum is 18, and the range is (−∞,18].

5A student claims the following two expressions are equal. Correct the error and find the zeros:

2x^2-8x+5\overset{?}{=}2(x-2)^2+5

$2(x-2)^2+5=2x^2-8x+13$

Expanding reveals the mismatch in the constant term.

$2x^2-8x+5=2\bigl((x-2)^2-4\bigr)+5=2(x-2)^2-3$

The correct vertex is (2,−3), not (2,5).

$(x-2)^2=\frac32\ \Longrightarrow\ x=2\pm\frac{\sqrt6}{2}$

The two exact zeros are real and irrational.

07 · Revisit

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