Analyze quadratic functions in vertex form

Alberta Grade 11 Mathematics

Relations and Functions

Locate the turning point before finding intercepts

Vertex form makes the turning point of a parabola visible without completing a new calculation. In Mathematics 20-1, this is useful for sketching a graph, identifying a maximum or minimum, and checking symmetry. The plotted curve in this lesson is calculated from its polynomial coefficients, so the graphic follows the worked algebra rather than a hand-drawn guess.

What you will learn

Read the vertex and direction from vertex form.
Find axis intercepts with substitution and factoring.
Check graph symmetry using two inputs equally distant from the vertex.

Read the vertex first

In

y=a(x-h)^2+k

, the vertex is

(h,k)

and the axis of symmetry is

x=h

. If a is positive, the parabola opens upward and k is its minimum y-value. If a is negative, it opens downward and k is its maximum.

A graphing window shows only part of an unlimited parabola. Use the equation, not the edge of the picture, when describing its full domain or range. The domain of a quadratic is all real numbers.

y=a(x-h)^2+k

Intercepts provide a check

For the y-intercept, substitute x = 0. For x-intercepts, set y = 0 and solve the resulting quadratic equation. There may be zero, one, or two real x-intercepts.

Points the same horizontal distance from the axis have the same y-value. Plot the vertex and at least one such matching pair to avoid an asymmetric sketch.

f(h-t)=f(h+t)

Worked example

Sketch a shifted parabola

Analyze

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

: find its vertex, axis, opening direction, intercepts, and range.

Graph of f(x) = (x − 2)² − 4

The displayed window shows the vertex and both x-intercepts; the parabola continues beyond the window.

Read the vertex
Here a = 1, h = 2, and k = −4. The graph opens upward, with axis x = 2.
$\text{vertex }$ (2,-4), x=2
Find intercepts
At x = 0, f(0) = 0. For x-intercepts, set the squared expression equal to 4.
0=(x-2)^2-4 \Rightarrow x=0 or x=4
State the range
An upward-opening parabola cannot go below its vertex.
$y≥-4$

Answer: Vertex

(2,-4)

; axis

x=2

; opens upward; x-intercepts

(0,0)

and

(4,0)

; y-intercept

(0,0)

; range

y\geq-4

.

Check:

f(0)=0

and

f(4)=0

, so the two x-intercepts are symmetric about

x=2

.

Common mistakes and how to avoid them

Reading h as −2 from

(x-2)^2

.

Correction: Vertex form uses

(x-h)^2

, so h = 2.

Treating the graph window as the whole domain.

Correction: The parabola continues indefinitely left and right.

Lesson summary

Vertex form directly reveals $(h,k)$ .
The sign of a sets the opening direction.
Substitution and symmetry check plotted intercepts.

Check your understanding

Question 1

What is the vertex of

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

?

$(−3,−1)$
$(3,−1)$
$(−3,1)$

Show answer and explanation

(−3,−1)

x+3=x-(-3)

, so h = −3 and k = −1.

Key terms

Vertex: The turning point of a parabola.
Axis of symmetry: A vertical line through the vertex that divides a parabola into matching halves.

Continue through Math 20-1

View the complete Math 20-1 Alberta Grade 11 Mathematics curriculum and lessons

About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Alberta Grade 11 Mathematics (Math 20-1), expectation RF3. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

Official curriculum reference

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RF3 · Analyze quadratic functions in vertex form