Math 2–3 Honors · Lesson Review · October 13, 2026

Graphing Quadratic Functions

Graph parabolas using vertices, symmetry, shifts, stretches, and reflections. Convert between standard and vertex forms and find an equation from a graph.

59-minute edited lesson9 chaptersCaptions + transcript

01 · Watch

Lesson video

VTT captions SRT captions

0:0001Equations, functions, and graphs 9:5602Build the parent parabola 18:3503Shifts and the square-number pattern 26:2204Vertex form and horizontal shifts 33:3505Vertical stretches 37:3506Compression and reflection 42:3407Factored form and completing the square 50:5808The standard-form vertex formula 57:1309Find an equation from a graph

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02 · Understand

Learning targets

Recognize a quadratic function and the shape of its graph.
Locate the vertex and axis of symmetry from vertex form.
Graph shifts, vertical stretches or compressions, and reflections of y=x².
Find the vertex from standard form and use a graph to determine an equation.

03 · Connect

Key ideas

The lesson builds the parent parabola from a table, transforms it, and connects vertex form with standard form and graph information.

01Quadratic functions make parabolas: $f(x)=ax^2+bx+c,\qquad a\ne0$ A quadratic function has degree two. Its graph is a parabola, unlike a linear function’s straight line. The domain is all real numbers.
02Read the vertex carefully: $f(x)=a(x-h)^2+k\quad\Longrightarrow\quad\text{vertex }(h,k),\ \text{axis }x=h$ The horizontal coordinate comes from making x−h zero. Thus x+3 means h=−3, while x−2 means h=2. The outside constant sets the vertical coordinate.
03Use symmetric offsets: $f(h+d)=f(h-d)=k+ad^2$ Choose offsets d=1, 2, and 3 from the vertex. The vertical changes are a, 4a, and 9a. Plot both matching points for each offset, then draw a smooth curve.
04Separate shifts from vertical scaling: $a>0:\ \text{opens up}\qquad a<0:\ \text{opens down}$ Changing h and k moves the vertex. Multiplying by a scales vertical distances from the vertex; a negative a also reflects the parent shape. Larger |a| appears narrower, while 0<|a|<1 appears wider.
05Find a standard-form vertex: $h=-\frac{b}{2a},\qquad k=f(h)$ Use the coefficients from ax²+bx+c to find the horizontal coordinate, then substitute that value into the original function. Completing the square gives the same vertex and makes the transformations visible.
06Factored form reveals x-intercepts: $f(x)=a(x-r_1)(x-r_2)$ When the roots are real, r1 and r2 give the x-intercepts. The teacher’s hypothetical factored example is a separate function; it is not a factorization of the standard-form example being graphed.
07Use the opening direction to state the range: $a>0:\ y\ge k\qquad a<0:\ y\le k$ An upward parabola has a minimum value k. A downward parabola has a maximum value k. For a nonzero quadratic coefficient, the graph continues without bound in its opening direction.

04 · Apply

Worked examples

These examples follow the methods developed in the recorded lesson.

Classroom example 1

Build the parent parabola

f(x)=x^2

Start with a table of squares.

f(0)=0,\quad f(\pm1)=1,\quad f(\pm2)=4,\quad f(\pm3)=9

Opposite inputs give equal outputs, so the graph is symmetric.

\text{vertex }(0,0),\quad\text{axis }x=0

Plot a smooth U-shaped parabola. Its domain is all real numbers and its range is y≥0.

Classroom example 2

Shift the vertex right and down

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

Compare with a(x−h)²+k: a=1, h=2, k=−3.

\text{vertex }(2,-3),\quad\text{axis }x=2

Move the parent vertex right 2 and down 3.

f(1)=f(3)=-2,\quad f(0)=f(4)=1

Place symmetric points one and two units from the vertex. The parabola opens upward and has range y≥−3.

Classroom example 3

Stretch, compress, and reflect

f(x)=2(x+1)^2-5

The vertex is (−1,−5), the axis is x=−1, and a=2 gives an upward vertical stretch.

f(-2)=f(0)=-3,\quad f(-3)=f(1)=3

Use vertical changes 2 and 8 at offsets 1 and 2.

g(x)=-\frac12(x+1)^2+3

This second function has vertex (−1,3), opens downward, and is vertically compressed.

g(-2)=g(0)=\frac52,\quad g(-3)=g(1)=1

Its range is y≤3. The negative sign changes direction; the factor 1/2 changes the scale.

Classroom illustration — hypothetical function

Use factors to locate x-intercepts

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

The teacher explicitly introduces this as a separate hypothetical function, not a factorization of x²−2x−5.

f(3)=f(-2)=0

The factors give x-intercepts (3,0) and (−2,0). Vertex form makes the vertex directly visible.

Classroom example 4

Complete the square to reveal the vertex

f(x)=x^2-2x-5

Use vertex form to graph a function originally written in standard form.

x^2-2x-5=(x^2-2x+1)-1-5

Add the completing-square term inside the expression and subtract it back so the function stays unchanged.

f(x)=(x-1)^2-6

The vertex is (1,−6) and the axis is x=1.

f(0)=f(2)=-5,\quad f(-1)=f(3)=-2

Graph symmetric points. The parabola opens upward and has range y≥−6.

Classroom example 5

Use the standard-form vertex formula

f(x)=x^2-6x+5

Identify a=1, b=−6, and c=5.

h=-\frac{-6}{2(1)}=3

The axis of symmetry is x=3.

k=f(3)=9-18+5=-4

The vertex is (3,−4).

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

Completing the square confirms the result. Symmetric points include (2,−3), (4,−3), (1,0), and (5,0).

Classroom example 6

Read an equation from a graph

\text{vertex }(-2,3)\quad\Longrightarrow\quad y=a(x+2)^2+3

Read the vertex first.

(-1,2):\quad 2-3=a(1)^2\quad\Longrightarrow\quad a=-1

The teacher reads a one-unit step right and a one-unit step down from the vertex. The vertical change gives a.

y=-(x+2)^2+3

The axis is x=−2 and the maximum value is 3.

(0,-1):\quad -1=4a+3\quad\Longrightarrow\quad a=-1

Added algebraic check: another plotted point independently confirms the same coefficient.

05 · Avoid

Common mistakes

Reading the inside sign directly as the vertex coordinate.

Find the input that makes the squared quantity zero: x+3 is zero at −3, so the horizontal coordinate is −3.

Treating a parabola like a straight line.

Use squared offsets 1, 4, and 9, multiplied by a. Equal horizontal steps do not give a constant slope.

Changing the function while completing the square.

When rewriting an expression, add and subtract the same amount. If there is an outside coefficient, account for its effect.

Giving only −b/(2a) as the vertex.

That formula gives h. Substitute h into the original function to find k, and write the vertex as the ordered pair (h,k).

Ignoring opening direction when stating the range.

An upward parabola has y≥k; a downward parabola has y≤k. Include the vertex value in the range.

06 · Check yourself

Try it first

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

1Find the vertex and axis, and give two symmetric points:

y=(x+4)^2-2

$\text{vertex }(-4,-2),\quad\text{axis }x=-4$

The parabola opens upward.

$y(-5)=y(-3)=-1$

Two matching points are (−5,−1) and (−3,−1). The range is y≥−2.

2Describe the transformations and graph symmetric points:

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

$a=3,\quad h=1,\quad k=2$

Shift right 1 and up 2, and stretch vertically by 3.

$\text{vertex }(1,2),\quad y(0)=y(2)=5$

The axis is x=1, the graph opens upward, and its range is y≥2.

3Find the vertex, opening direction, range, and matching points:

y=-2(x+2)^2+5

$\text{vertex }(-2,5),\quad\text{axis }x=-2$

The negative coefficient gives a downward opening, so the range is y≤5.

$y(-3)=y(-1)=3$

At one-unit offsets, the output is 2 below the vertex.

4Rewrite in vertex form and identify the vertex:

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

$2(x^2-4x)+3=2(x^2-4x+4)-8+3$

Complete the square inside the factored expression. Adding 4 inside adds 8 overall, so subtract 8 outside.

$f(x)=2(x-2)^2-5$

The vertex is (2,−5), the axis is x=2, and the range is y≥−5.

5Find the equation of a parabola with vertex (3,−1) passing through (5,7):

y=a(x-3)^2-1

$7=a(5-3)^2-1\quad\Longrightarrow\quad8=4a$

Substitute the point that is not the vertex.

$a=2\quad\Longrightarrow\quad y=2(x-3)^2-1$

It opens upward, has axis x=3, and has range y≥−1.

07 · Revisit

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