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A. Quadratic Functions · Quadratic Functions

Sketching Quadratic Functions by Transforming f(x) = x²

A2.6 A2.5About 65 minutes
Learning goals
  • I can sketch g(x) = a(x − h)² + k by transforming the key points of f(x) = x².
  • I can sketch using the vertex, the step pattern and symmetry.
  • I can compare several transformed parabolas on one grid.
  • I can read the zeros, y-intercept and range from my sketch.

Words to know

sketch
A quick graph that shows the key features (vertex, direction, shape, a few exact points), not a careful plot of many points.
key points
Easy points used to sketch: for y = x², (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4).
step pattern
From the vertex of y = a(x − h)² + k, the rises a, 3a, 5a for each unit step sideways.
axis of symmetry
The vertical line x = h through the vertex; the parabola is its own mirror image across it.

Sketch it in under a minute

A sketch is not a careful plot of twenty points. It is a quick, accurate picture that shows the important features: the vertex, the direction of opening, the shape, and a few exact points. Engineers, scientists and economists sketch functions all the time to see what is going on before they calculate.

In the last lesson you learned what a, h and k do. Now you will use them to sketch g(x) = a(x − h)² + k by transforming the graph of f(x) = x².

Method 1: transform the key points of f(x) = x²

The five key points of the parent are (−2, 4), (−1, 1), (0, 0), (1, 1) and (2, 4).

  1. Stretch, compress or reflect: multiply each y-coordinate by a.
  2. Translate: add h to each x-coordinate and k to each y-coordinate.
  3. Plot the five new points and join them with a smooth U-shaped curve (not straight segments). Add arrows or extend the curve to show it continues.

This is the mapping rule (x, y) → (x + h, ay + k) done in two stages.

xy−3036−8−4041. f(x) = x²xy−3036−8−4042. multiply y by −2xy−3036−8−4043. move right 3, up 4
Sketching g(x) = −2(x − 3)² + 4 in three stages. Multiplying every y-coordinate by −2 flips and stretches the parent. Then adding 3 to every x and 4 to every y moves the vertex from (0, 0) to (3, 4).
Worked example

Example 1: Sketch g(x) = −2(x − 3)² + 4

Sketch g(x) = −2(x − 3)² + 4 by transforming f(x) = x². State the vertex, axis, domain and range.

  1. a = −2, h = 3, k = 4.
  2. Multiply y by −2: (−2, 4) → (−2, −8); (−1, 1) → (−1, −2); (0, 0) stays; (1, 1) → (1, −2); (2, 4) → (2, −8).
  3. Add 3 to x and 4 to y: (−2, −8) → (1, −4); (−1, −2) → (2, 2); (0, 0) → (3, 4); (1, −2) → (4, 2); (2, −8) → (5, −4).
  4. Plot these and draw a smooth curve opening down. Vertex (3, 4), axis x = 3.
  5. D = {x ∈ ℝ}; R = {y ∈ ℝ | y ≤ 4}.

The sketch passes through (1, −4), (2, 2), (3, 4), (4, 2) and (5, −4). Vertex (3, 4), axis x = 3, R = {y ∈ ℝ | y ≤ 4}.

Method 2: the step pattern from the vertex

For y = x², moving 1, 2, 3 units away from the vertex takes you up 1, 4, 9 units. Step by step, that is up 1, then up 3, then up 5.

For g(x) = a(x − h)² + k, multiply the pattern by a:

  1. Plot the vertex (h, k).
  2. From the vertex, go 1 right and up a, then 1 right and up 3a, then 1 right and up 5a. ("Up" a negative amount means down.)
  3. Use symmetry: copy each point to the left of the axis x = h.

This method is fast and works well when a is a simple number like 2, 0.5 or −3.

xy−7−6−5−4−3−2−10123−4−3−2−10123456+0.5+1.5+2.5vertex (−2, −3)g(x) = 0.5(x + 2)² − 3steps: 0.5 × (1, 3, 5)
For g(x) = 0.5(x + 2)² − 3, start at the vertex (−2, −3). Each step right raises the curve by 0.5, then 1.5, then 2.5 (half of 1, 3 and 5). The points on the left are mirror images across the axis x = −2.
Worked example

Example 2: Sketch g(x) = 0.5(x + 2)² − 3

Sketch g(x) = 0.5(x + 2)² − 3 using the step pattern. State the y-intercept, and estimate the zeros from the sketch.

  1. Vertex (−2, −3). a = 0.5, so the steps are 0.5 × 1 = 0.5, 0.5 × 3 = 1.5 and 0.5 × 5 = 2.5.
  2. Right side: (−1, −2.5), (0, −1), (1, 1.5). Left side by symmetry: (−3, −2.5), (−4, −1), (−5, 1.5).
  3. y-intercept: g(0) = 0.5(2)² − 3 = 2 − 3 = −1. (It matches the point (0, −1).)
  4. The curve crosses the x-axis between x = 0 and x = 1 (from −1 up to 1.5), and between x = −5 and x = −4.
  5. Estimate: zeros near x ≈ 0.4 and x ≈ −4.4. (Exactly, 0.5(x + 2)² = 3 gives x = −2 ± √6 ≈ 0.45 and −4.45.)

Vertex (−2, −3), y-intercept −1, zeros about 0.4 and −4.4.

Worked example

Example 3: Comparing sketches on one grid

On one set of axes, sketch f(x) = x², g(x) = (x − 1)² − 4 and h(x) = −(x − 1)² + 4. Describe how g and h are related.

  1. f: the parent, vertex (0, 0).
  2. g: vertex (1, −4), a = 1, steps up 1, 3, 5: (2, −3), (3, 0), (4, 5) and mirror points (0, −3), (−1, 0), (−2, 5). Zeros −1 and 3.
  3. h: vertex (1, 4), a = −1, steps down 1, 3, 5: (2, 3), (3, 0), (4, −5) and (0, 3), (−1, 0), (−2, −5). Zeros −1 and 3.
  4. g and h have the same axis x = 1 and the same zeros −1 and 3. h is g reflected in the x-axis: h(x) = −g(x).

g and h are reflections of each other in the x-axis. They share the axis x = 1 and the zeros −1 and 3.

Quality checklist for a sketch

  • Vertex plotted and labelled.
  • At least two points on each side, placed symmetrically.
  • A smooth curve, not a V shape, that keeps going past the last points.
  • The y-intercept is right (check g(0)).
  • The direction matches the sign of a.
Just for fun

Why don't parabolas ever get lost on a road trip?

They always know exactly where they turn around: at the vertex.

Hands-on activity

The 60-second sketch challenge

You need: a partner, graph paper (or plain paper with hand-drawn axes), a pencil, a timer (a phone works), and a set of cards or slips of paper. Desmos or a graphing calculator is optional for checking.

  1. Your partner writes six equations of the form g(x) = a(x − h)² + k on slips, using values like a = 2, −1, 0.5, −3 and small whole numbers for h and k.
  2. Draw a slip. Start the timer. Sketch the function in 60 seconds or less, using either method.
  3. Your partner checks the sketch: vertex, direction, and two points each side. Check with a table or Desmos if you are unsure.
  4. Swap roles. Then try the reverse: your partner sketches a parabola with a clear vertex and one extra point, and you find its equation.

What to look for: a correct vertex, the right direction, correct steps (a, 3a, 5a), symmetry, and a smooth curve. For the reverse task, the equation should use the vertex for h and k and the extra point to solve for a.

Your turn

1

List the vertex and four more points to sketch g(x) = (x − 2)² + 1.

2

List the vertex and four more points to sketch g(x) = −(x + 1)² + 3.

3

Transform the five key points of y = x² for g(x) = 3x² − 2.

4

List five points to sketch g(x) = 0.5(x − 4)².

5

What is the step pattern for a = −2?

6

Sketch g(x) = 2(x + 3)² − 8. Use your points to state the zeros and the y-intercept.

More practice: warm-up

1

List the five key points of f(x) = x².

2

What is the first point to plot for g(x) = (x − 5)² + 2?

3

For a = 3, how much higher than the vertex are the points 1 unit to its left and right?

4

What is the step pattern when a = 1?

5

State the y-intercept of g(x) = x² + 4.

More practice: on your own

1

Sketch g(x) = −0.5(x − 2)² + 8. State the zeros and the range.

2

Sketch g(x) = 4(x − 1)² − 1 and find its zeros. (Hint: try x = 0.5.)

3

On one grid, sketch f(x) = x², g(x) = x² − 4 and h(x) = (x − 4)². How many zeros does each have?

4

A sketch has vertex (−1, 2) and passes through (0, 5). Find its equation.

5

Sketch g(x) = −(1/3)(x + 3)² + 3. Use x = 0 and x = −6 to find the zeros.

Challenge

1

Sketch g(x) = 2(x − 1.5)² − 4.5. Show that the graph passes through the origin, and find the other zero without solving an equation.

2

g(x) = (x − 2)² + k and h(x) = −(x − 2)² + 6. For what value of k do the graphs touch at exactly one point? What happens for other k?

3

A parabola has vertex (3, −2) and passes through (5, 6). Find its equation, then use the step pattern to find its zeros.

Check your understanding

Quiz

1.

What are the vertex and direction of g(x) = 3(x − 4)² − 1?

2.

For a = 2, the step pattern from the vertex is…

3.

Which point is on the graph of g(x) = −(x + 2)² + 5?

4.

For g(x) = 0.5x² + 1, what is g(2)?

5.

How many zeros does g(x) = 2(x − 1)² + 3 have?

6.

When transforming the key points of y = x², which step comes before adding k?

For parents and teachers

Big idea: expectation A2.6 asks students to sketch g(x) = a(x − h)² + k by applying one or more transformations to the graph of f(x) = x².

How to help: ask your teen to talk through each sketch: "The vertex is…, it opens…, the steps are…." Time a few sketches; speed comes with practice. Checking one point in the equation catches most errors.

Common difficulties:

  • Taking steps from the origin instead of from the vertex.
  • Forgetting to multiply the step pattern by a (using 1, 3, 5 for every parabola).
  • Joining points with straight lines or drawing a V at the vertex.
  • Plotting points on only one side of the axis.