Quadratic and Polynomial Functions
Finding the Vertex and Graphing a Quadratic
Two routes to the turning point, the four questions a complete graph answers, and a worked sketch from start to finish.
What this page covers
- Find a vertex by completing the square
- Find a vertex using x = -b2a
- Find the intercepts of a parabola
- Assemble a complete labelled sketch
Two routes to the vertex
Complete the square
Gives the full vertex form a(x - h)2 + k, so both coordinates appear at once. More work, more information.
Use x = -b2a
Gives the x coordinate directly; substitute back for y. Faster when only the turning point is wanted.
The formula is completing the square done once and remembered. It is the same -b2a that sits outside the radical in the quadratic formula, which is why the roots are always symmetric about the vertex.
Completing the square, worked
y = 2x2 - 12x + 4
Minimum -14 at x = 3. Note the compensation: adding 9 inside a bracket multiplied by 2 adds 18 overall, so 18 must be subtracted outside.
-b2a = --124 = 3 ✓.
y = 12x2 + 3x
Minimum -92 at x = -3. The compensation is scaled by 12, not by 1.
Whatever is added inside the bracket is multiplied by the factor outside it. With a factor of 2 the compensation is 2a2; with 12 it is 12a2. Using a2 regardless is the standard error in the non-monic case.
y = 2x2 - 5x + 3
Minimum -18 at x = 54, which the source marks on its sketch as (54, -18).
Where the parabola meets the axes
The typed notes set out four questions a complete graph should answer, and they make a good checklist.
- Where does it cut the x-axis? Set y = 0 and solve. These are the roots or zeros.
- Where does it cut the y-axis? Set x = 0; the answer is c.
- Where is the vertex? Complete the square, or use -b2a.
- What is the overall shape? Upward if a > 0, downward if a < 0.
Finding the roots — the source's case
Where does y = 2x2 - 5x + 3 cut the x-axis?
The source's line reads 25 ± √14 = 264, 244. The 25 has been carried down from under the radical; outside it sits -b = 5. Substituting confirms the corrected roots: 2(94) - 152 + 3 = 0 and 2 - 5 + 3 = 0.
The roots should average to the axis: 12(32 + 1) = 54, which matches the vertex found above. The uncorrected roots average to 6.25, which does not — the check catches the error at once.
A complete worked sketch
Sketch y = x2 - 8x + 6
This is the typed notes' running example.
| Question | Working | Answer |
|---|---|---|
| Shape | a = 1 > 0 | Opens upward |
| y-intercept | x = 0 | (0, 6) |
| Vertex x | -b2a = 82 | x = 4 |
| Vertex y | 16 - 32 + 6 | y = -10 |
| Roots | x = 8 ± √64 - 242 = 4 ± √10 | ≈ 0.84 and ≈ 7.16 |
Completing the square confirms the vertex directly:
The two roots and the axis of symmetry between them. √10 ≈ 3.16, so the roots lie 3.16 either side of 4.
The roots average to 4, matching the axis. The vertex lies below the axis and the parabola opens upward, so two roots are expected — and two were found.
How many roots to expect
| Opens | Vertex y | Roots | Discriminant |
|---|---|---|---|
| Up | Below the axis | Two | Positive |
| Up | On the axis | One (repeated) | Zero |
| Up | Above the axis | None (real) | Negative |
| Down | Above the axis | Two | Positive |
| Down | On the axis | One (repeated) | Zero |
| Down | Below the axis | None (real) | Negative |
This gives a check that costs nothing. If the vertex is computed as (4, -10) with an upward parabola, two roots must exist; finding none would indicate an arithmetic error rather than a genuine absence.
Common mistakes
| Mistake | Correct | Check |
|---|---|---|
| Not scaling the compensation when a ≠ 1 | Multiply it by a | Expand back and compare |
| x = b2a | x = -b2a | The roots must average to it |
| Reading c as the vertex y | c is the y-intercept | They coincide only when b = 0 |
| Carrying -b under the radical | -b is outside | Check the roots average to -b2a |
| Plotting many points instead of the key ones | Vertex plus intercepts is enough | Four questions, four answers |
| Expecting two roots when the vertex is above the axis | None, for an upward parabola | Check the discriminant |
Frequently asked questions
Which method for the vertex is better?
x = -b2a is faster when only the vertex is wanted. Completing the square is better when the whole vertex form is needed, and it is where the formula comes from.
Where does x = -b2a come from?
From completing the square in general: ax2 + bx + c becomes a(x + b2a)2 + …, and the bracket vanishes at x = -b2a. The source states the rule as minimum is always at x = -b2a.
What are the four questions a graph should answer?
The typed notes list them: where does it cut the x-axis, where the y-axis, where is the vertex, and what is the overall shape.
Do I need to plot many points?
No. The typed notes say it is acceptable to plot the key points and draw the curve through them. The vertex and the intercepts, plus the direction of opening, are enough.
Source. Handwritten teaching notes, Week 6, pages 3-4, with the typed supplementary notes on solving and graphing quadratic functions.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.