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If the sign inside the bracket is confusing, begin with the first four chapters. If you know the turning point but mix it up with intercepts, jump to the sections on the x-axis and y-axis. For quadratic sketches, read the symmetry, reflection and completed-square chapters before trying the practice clinic.
For y = (x − 3)² + 2, the turning point is (3,2), not (−3,2). The squared bracket reaches its minimum value of zero when x − 3 = 0, so x must be three. The outside constant then gives y = 2. This is the cleanest way to read the location without guessing from the printed minus sign.
The same expression shows more: the parabola opens upward, has axis of symmetry x = 3, passes through (0,11) on the y-axis, and has no real x-intercepts because its value can never fall below two. Understanding those features is more useful than memorising vertex coordinates in isolation.
The 2027 SEC G3 Mathematics K310 syllabus explicitly covers quadratic graphs, their extrema and symmetry, including sketches in completed-square form. This guide starts with those ideas and includes labelled extensions. Use the official syllabus and the student’s school programme to distinguish required work from deeper exploration.
- Quick read
- 1. Read the completed-square form before plotting
- 2. Why the sign inside a bracket appears reversed
- 3. The constant outside the square is the vertical position
- 4. Full worked solution: intercepts, axis and minimum
- 5. A minimum is a y-value; a turning point is a coordinate
- 6. The axis of symmetry divides matching outputs
- 7. Positive squared coefficient gives a minimum
- 8. A negative squared coefficient gives a maximum
- 9. The value of p is found from the zero of the bracket
- 10. Find the y-intercept by setting x to zero
- 11. Find x-intercepts by setting y to zero
- 12. A turning point may lie exactly on the x-axis
- 13. Completed-square form and factorised form answer different questions
- 14. Completing the square explains the turning point in expanded form
- 15. The sign of the linear coefficient is not the turning-point coordinate
- 16. Distinguish maximum from the highest visible data point
- 17. A quadratic graph is symmetrical, not linear
- 18. A reflected parabola preserves its horizontal centre
- 19. A horizontal translation is not the same as a vertical translation
- 20. The graph can have no real x-intercepts
- 21. The graph can have exactly two x-intercepts
- 22. Interpret the turning point in a contextual model
- 23. A graph drawn not to scale cannot decide the exact vertex
- 24. Check the vertex with a short numerical table
- 25. Diagnose sign errors before adding more practice
- 26. Connect the method to the official G3 Mathematics syllabus
- 27. A final routine for vertex-form questions
- 21 solved quadratic checks
- Frequently asked questions
In y = a(x − p)² + q with nonzero a, the bracket vanishes at x = p, giving turning point (p,q). When a is positive the point is a minimum; when a is negative it is a maximum. The symmetry axis is x = p. Find the y-intercept by setting x = 0 and x-intercepts by setting y = 0; neither is automatically the turning point.
For y = (x − 3)² + 2, the minimum value is two at x = three. The vertex is (3,2), symmetry axis x = 3, y-intercept (0,11), and there are no real x-intercepts. Check equal outputs at equal distances from x = 3, such as x = 2 and x = 4.
The expression y = (x − 3)² + 2 is a quadratic written in completed-square or vertex form. The bracketed square can never be negative for a real x, so the smallest possible contribution from the square is zero. It becomes zero when x − 3 = 0, giving x = 3. At that x-value, y is two.
Therefore the turning point is (3, 2). The number three is not chosen because there is a minus sign printed in the bracket; it is chosen by solving the bracket equation. This distinction protects learners from reporting the incorrect point (−3, 2).
Reading the algebra before sketching also reveals that the parabola opens upward because the squared term has a positive coefficient. The turning point is a minimum.
Always write ‘make the squared bracket zero’ before deciding the horizontal coordinate. The printed sign alone is not a reliable shortcut.
A horizontal shift occurs inside the input of a function. The expression x − 3 becomes zero at x = 3, so the centre of the squared shape is three units to the right of the origin. A learner who reads the minus as ‘move left’ has confused the operation inside the formula with the resulting location on the graph.
Compare y = x², whose minimum occurs at x = 0, with y = (x − 3)², whose minimum occurs at x = 3. Their shapes are identical but the latter is shifted right three units. For y = (x + 3)², the zero of the bracket occurs at x = −3 and the graph shifts left.
The rule follows from solving a simple equation rather than memorising a visual slogan about opposite signs.
When uncertain, substitute x = 3 and x = −3. Only x = 3 makes (x − 3)² equal zero.
In y = (x − 3)² + 2, the outside plus two is added after the square has been calculated. Because the squared term reaches zero, this constant gives the lowest y-value. It translates the basic parabola upward without changing the x-coordinate at which its minimum occurs.
The minimum of y = (x − 3)² is zero at x = 3. The minimum of y = (x − 3)² + 2 is two at the same x. Replacing plus two with minus four produces a minimum of negative four at x = 3.
The vertical shift does not depend on the sign of x. It applies to every point on the curve because the same number is added to every output.
Underline terms outside the squared bracket separately from those inside. The two positions correspond to different geometric movements.
Consider y = (x − 3)² + 2. First identify the minimum: the square is zero when x = 3, so the turning point is (3, 2). The vertical line x = 3 is the axis of symmetry. These conclusions are exact and do not require calculating many points.
To find the y-intercept, put x = 0. Then y = (−3)² + 2 = 11, so the curve crosses the y-axis at (0,11). For x-intercepts set y = 0; the equation (x − 3)² + 2 = 0 would require a square to equal negative two, impossible over the real numbers. Therefore there is no x-intercept.
The worked example answers three different questions: where the graph turns, where it meets the y-axis, and whether it meets the x-axis. These are not synonyms.
Verify the shape by computing symmetric values at x = 2 and x = 4. Both give y = 3, one unit above the minimum.
| Feature | Result | Method |
|---|---|---|
| Vertex | (3,2) | (x − 3)² = 0 at x = 3 |
| Type | Minimum | Positive coefficient of square |
| Axis | x = 3 | Through vertex |
| y-intercept | (0,11) | Set x = 0 |
| x-intercepts | None in real numbers | Square + 2 cannot equal zero |
Questions may ask for the minimum value of a function, the x-value at which the minimum occurs, or the coordinates of the turning point. The answers are related but have different formats. A student who writes ‘(3, 2)’ when only the minimum value is required has supplied extra information; writing ‘2’ when coordinates are required is incomplete.
For y = (x − 3)² + 2, the minimum value is 2, it occurs when x = 3, and the turning point is the ordered pair (3,2). The axis of symmetry is the entire vertical line x = 3.
The distinction becomes especially important in graph interpretation and word problems, where x and y may have different physical units.
Underline the requested noun: coordinate, value, axis, input or intercept. Use the right mathematical object and units in the final line.
A quadratic in vertex form y = (x − p)² + q is symmetrical about x = p. Inputs equally far to the left and right of p produce the same squared distance from the centre. This property provides an independent numerical test for the identified turning point.
For y = (x − 3)² + 2, x = 1 and x = 5 are two units from the axis. Both give y = 4 + 2 = 6. Likewise, x = 2 and x = 4 both produce y = 3. These matched pairs flank x = 3.
A table that has unequal outputs for such a symmetric pair signals a substitution or arithmetic mistake, not a genuinely asymmetrical parabola.
Choose pairs p − d and p + d, substitute and compare. The common output should be d² + q.
In y = a(x − p)² + q, a positive nonzero coefficient multiplies a nonnegative squared term, so the contribution is also nonnegative. The smallest value remains q at x = p. The size of a changes the steepness or vertical stretch of the graph but not the location of that turning point.
For y = 2(x − 3)² + 2, the turning point is still (3,2). At x = 2 the function gives 4 rather than 3, because the squared distance of one is multiplied by two. The graph rises more quickly away from its centre.
The coefficient a belongs outside the square. Confusing a change in a with a horizontal shift would produce a wrong sketch even if the vertex coordinates happened to be correct.
Read sign and magnitude separately. Positive tells the orientation, while magnitude controls how strongly the curve grows away from the axis.
If the coefficient of the squared bracket is negative, the squared term contributes zero or a negative amount. The function therefore reaches its greatest value q when the bracket is zero. The turning point remains at x = p, but it becomes a maximum instead of a minimum.
For y = −(x − 3)² + 2, the turning point is (3,2) and maximum value is 2. At x = 2 or x = 4, y = 1. At x = 1 or x = 5, y = −2. These points form a downward-opening parabola.
A downward-opening graph can have positive outputs near its maximum and negative outputs farther away. Orientation alone does not determine every y-value.
Look at the sign directly before the squared bracket, then name maximum or minimum. Do not let the inside minus sign make that decision.
In the expression (x − p)², the number p can itself be positive, negative, zero or fractional. The universal method is to solve x − p = 0. This avoids confusing cases where the inside of the bracket contains a plus sign because p happens to be negative.
For y = (x + 5)² − 1, rewrite the bracket as x − (−5). It vanishes at x = −5, so the turning point is (−5, −1). For y = (x − 1/2)² + 4, the minimum occurs at x = 1/2.
The method also makes decimals straightforward. There is no special sign rule to learn for fractional shifts.
Ignore superficial pattern matching and solve the one-step equation. It gives the axis correctly for all real values of p.
The y-intercept is where the graph meets the vertical axis, so x must equal zero. A turning point usually has a different x-coordinate, and a squared expression may produce a large y-intercept despite a small minimum. Do not confuse the constant q in vertex form with the y-intercept unless p equals zero.
For y = (x − 3)² + 2, setting x = 0 gives 11. The y-intercept is (0,11), not (0,2). The number two describes the vertical coordinate of the turning point, not automatically the vertical-axis crossing.
This contrasts with expanded form y = ax² + bx + c, where c is the y-intercept because x = 0 removes the other terms.
Substitute zero explicitly whenever an intercept is requested. The correct question is ‘which axis am I on?’ rather than ‘which constant can I see?’
An x-intercept occurs when the curve meets the horizontal axis, where y = 0. Setting the squared expression equal to zero or to a negative constant can reveal whether real x-intercepts exist. The turning point can sit above, on or below the x-axis, which changes the number of intersections.
For y = (x − 3)² − 4, set zero equal to the expression: (x − 3)² = 4. Thus x − 3 = 2 or −2, so x = 5 or x = 1. The x-intercepts are (5,0) and (1,0), while the turning point is (3,−4).
Listing only x = 1 and x = 5 is acceptable when the question asks for the solutions to y = 0, but intercept coordinates should include y = 0.
Check the two roots are symmetric about x = 3. Their midpoint is three, supporting the turning-point calculation.
When the outside constant q equals zero in y = (x − p)², the minimum value is zero, so the curve touches the x-axis at its turning point. This is one distinct x-intercept, even though the algebraic factor x − p appears twice when the square is expanded.
For y = (x − 3)², the only real solution of y = 0 is x = 3. The graph descends to (3,0) and then rises again, rather than crossing through the horizontal axis at that point.
The statement concerns a quadratic with positive coefficient. A corresponding downward-opening negative square also touches its horizontal axis at the vertex if q is zero.
Distinguish one distinct intercept from multiplicity of an algebraic root. The point is one coordinate, not two separate physical crossings.
SECTION 15 OF 31
13. Completed-square form and factorised form answer different questions
One quadratic can be written in multiple equivalent forms. Completed-square form exposes the turning point, while factorised form can expose x-intercepts. Expanded form displays the coefficients. A capable learner chooses the form that reveals the requested feature rather than repeatedly carrying out unnecessary transformations.
For y = (x − 3)² − 4, expansion gives x² − 6x + 5 and factorisation gives (x − 1)(x − 5). The first form shows vertex (3,−4), while the last shows zeros at one and five. All are exactly the same quadratic function.
The forms are connected by valid algebra and should agree at every input. Checking a single value can detect some mistakes, but equality of forms is established by expansion or factorisation.
Ask which feature the question seeks. Use vertex form for extrema and factorised form for roots when the factors are convenient.
SECTION 16 OF 31
14. Completing the square explains the turning point in expanded form
A quadratic may be supplied as x² − 6x + 11 rather than in vertex form. Completing the square rewrites it as (x − 3)² + 2. This allows students to locate the minimum without estimating its position from a drawn curve. The method is grounded in the identity (x − 3)² = x² − 6x + 9.
Start with x² − 6x + 11. Add and subtract nine by writing (x² − 6x + 9) + 2, then group the first three terms as (x − 3)² + 2. The turning point is (3,2).
The algebra preserves the function at every x because the plus-nine and minus-nine adjustment was balanced. No value was changed by the rewrite.
Verify the expansion of (x − 3)² + 2 before accepting the turning-point conclusion. A missing cross term causes an incorrect centre.
SECTION 17 OF 31
15. The sign of the linear coefficient is not the turning-point coordinate
In expanded form y = x² + bx + c, the location of the minimum is not simply b or −b. With unit leading coefficient, completing the square shows that the x-coordinate is −b/2. This is a useful extension for connecting coefficients to vertex form, but careless reading of the printed sign can double the actual shift.
For y = x² − 6x + 11, b = −6, and −b/2 = 3. The vertex-form bracket is x − 3, not x + 6 or x − 6. The coefficient −6 came from the cross term −2 × 3 × x.
The general quadratic formula for the vertex position uses −b/(2a) when a is nonzero. Learners should not generalise the unit-leading-coefficient shortcut without that condition.
Where available, confirm the vertex position by completing the square rather than relying on an unexamined formula.
A plotted graph or table may only show a handful of sample points. The highest or lowest value among those samples is not necessarily the true extremum of the underlying continuous quadratic. Vertex form gives an exact turning point from the algebra, even when the sample grid omits the required x-coordinate.
For y = (x − 2.5)² + 1, a table using only integer x-values gives y = 1.25 at x = 2 and x = 3. But the true minimum is 1 at x = 2.5, which lies between the sampled inputs.
Interpolation by eye may give a reasonable estimate but should not replace an exact algebraic conclusion when the function is known.
Check whether the turning-point x-coordinate lies between table entries. If it does, the table minimum may overstate the true minimum.
A quadratic in vertex form curves because the displacement from its centre is squared. Equal steps away from the centre produce increasingly larger vertical changes. Joining a few plotted points with straight segments can be a rough sketch, but the actual function is not a straight line with a single constant gradient.
For y = (x − 3)² + 2, at x = 3,4,5,6 the outputs are 2,3,6,11. The successive changes are +1,+3,+5 rather than a constant change. On the other side of the turning point, the same pattern reverses.
The symmetry and changing rates are both algebraic consequences of the square. They distinguish a parabola from a straight-line graph.
Plot enough points on both sides of the axis and draw a smooth curve through them, respecting the vertex and symmetry.
Changing the sign of the whole squared contribution reflects the graph vertically relative to its shifted base line. It does not move the turning point left or right. This provides another way to separate the coefficient outside the square from the expression inside it.
Compare y = (x − 3)² + 2 and y = −(x − 3)² + 2. Both have a vertex at (3,2). One has a minimum there, while the other has a maximum. At x = 2, the outputs are 3 and 1 respectively.
The graphs have the same vertical axis of symmetry. Reflecting orientation does not create a new value of p.
Make a small table for the same x-values in both functions and compare changes about the common horizontal line y = 2.
SECTION 21 OF 31
19. A horizontal translation is not the same as a vertical translation
Inside changes to x move the graph horizontally; outside additions move it vertically. Students can remember the distinction more reliably by thinking about what must be zero for the squared term to vanish and then what remains as the output. This works for any permitted real shift values.
In y = (x − 4)² + 7, the bracket is zero at x = 4 and the output is seven, giving turning point (4,7). In y = (x + 7)² − 4, the bracket is zero at x = −7 and the output is −4, giving (−7,−4).
The order of the printed signs changes their geometric roles because they appear in different places in the formula.
Separate each expression into a squared bracket and an outside constant. That visual split often resolves a sign confusion quickly.
A quadratic such as y = (x − 3)² + 2 stays at or above two, so it cannot equal zero. This is a direct argument from nonnegativity, not a guess from a graph drawn too high on the page. A student’s conclusion about roots should follow the function and its allowed domain.
To solve (x − 3)² + 2 = 0 over the reals, rearrange to (x − 3)² = −2. No real squared number is negative, so there is no real solution. The curve never meets the x-axis.
Complex-number solutions exist in an extended number system, but that is separate from asking for real x-intercepts on a Cartesian graph.
State ‘no real x-intercepts’ rather than ‘no turning point’. The turning point still exists at (3,2).
If the minimum of an upward-opening unit quadratic lies below zero, the curve can cross the x-axis at two locations symmetrical about the axis. The same happens for a downward-opening quadratic whose maximum lies above zero. Solve the squared equation to obtain both positions and check that the values are real.
For y = (x − 3)² − 9, set y = 0: (x − 3)² = 9. Hence x = 0 or x = 6. The axis is x = 3, halfway between them, while the turning point is (3,−9).
The existence of two roots does not alter the vertex coordinates. The roots are points on the x-axis, not the minimum itself.
Use symmetry to check the root pair: its mean should match the axis position for a quadratic with two real roots.
When a quadratic represents a quantity varying with x, the turning point can describe a minimum cost, maximum height or other extremum. Its significance depends on the meaning and domain of x, which may be restricted to times or lengths that make sense. The algebra can identify a turning point outside the permitted domain, in which case it is not the relevant practical optimum.
In a fictional mathematical model h(t) = −(t − 3)² + 10 for time 0 ≤ t ≤ 6, the height reaches a maximum ten units at t = 3. If the model were used only for times 4 ≤ t ≤ 6, its maximum on that restricted interval would occur at t = 4, not at the excluded t = 3.
Context and domain therefore matter after the algebraic turning point has been found. A graph’s global vertex and a model’s allowed optimum need not coincide.
Write the permitted interval and the relevant quantity before reporting a maximum or minimum in a word problem.
Textbook illustrations often prioritise readability. A printed parabola may appear to turn slightly left of its true algebraic coordinate because of drawing resolution, axes or line thickness. In a formula-based question, exact vertex information comes from the expression and stated conditions, not from estimating a pixel position.
For y = (x − 3)² + 2, a rough sketch could seem to turn near x = 2.8, but the exact bracket calculation still gives x = 3. If the question gives only a graph with scale information, a graphical estimate may instead be the intended method.
Distinguish an algebraically exact answer from an estimate read from data or a diagram. They have different levels of precision.
Use labelled axes and values where a numerical estimate is required, but prefer the algebraic form when it is supplied.
A few carefully chosen values can verify a turning-point reading. Evaluate at the claimed centre and equal offsets on both sides. The central output should be the extremum, and the offsets should give matching heights for the simple translated-square family. This is a useful independent check against sign inversion.
For y = (x − 3)² + 2, evaluate x = 2,3,4 to get 3,2,3. The smallest value is at x = 3 and the outside two is recovered exactly. If you mistakenly choose x = −3, you would obtain 38, not a minimum.
A table verifies the result but should not be the sole justification when the problem requests an algebraic explanation of the vertex.
After writing the proposed turning point, test one input to either side of its x-coordinate before moving on.
When a student places the vertex at (−3,2) for y = (x − 3)² + 2, the error is likely a failure to solve the zero-bracket condition. Another student may find the vertex correctly but call it a maximum because the outside constant is positive. These are different conceptual weaknesses that require targeted correction.
For y = −(x + 4)² − 1, setting x + 4 = 0 gives x = −4, and the negative leading sign makes y = −1 a maximum. A student who says (4,−1) needs a horizontal-sign repair; one who says minimum needs an orientation repair.
Generic warnings to ‘be careful’ do not identify which part of the formula was misread. Separate the two decisions: location and type of extremum.
Have the learner state the reason for x, the reason for y and the reason for maximum or minimum independently.
The 2027 SEC G3 Mathematics K310 syllabus includes graphs of quadratic functions, maxima and minima, symmetry and sketches in forms such as y = (x − p)² + q and y = −(x − p)² + q. The ability to read the turning point follows directly from the squared expression’s nonnegative nature.
Related G3 guides explain why a quadratic’s x-intercept is not its turning point and how other forms reveal roots. The How Mathematics Works hub provides connected secondary Mathematics reading, while eduKate’s Clementi teaching reference emphasises diagnosing the first reasoning error rather than only drilling steps.
More advanced coefficients and complex solutions can be treated as extensions. Distinguish those from the exact school assessment scope, using the official subject syllabus as the controlling reference.
A student who understands the zero-bracket explanation has a transferable method for unfamiliar sign variations and shifts.
Read y = a(x − p)² + q as a structured mathematical object. Solve the bracket equation x − p = 0 to obtain the horizontal coordinate. Evaluate the output at that x to obtain q. Determine whether the turning point is a maximum or minimum from the sign of a. Then answer intercept or graph questions separately.
For y = −(x + 2)² + 5, the bracket vanishes at x = −2, giving turning point (−2,5). Because a is negative, it is a maximum. The axis of symmetry is x = −2. To find the y-intercept, put x = 0, giving y = −4 + 5 = 1.
Notice how the maximum, axis and y-intercept are related without being the same thing. A well-structured final answer labels each feature clearly.
Finish by checking two symmetric points, such as x = −3 and x = −1, which both give y = 4.
Try each question from the given function before uncovering the answer. Identify the zero of the squared bracket, its resulting y-coordinate and the sign of the coefficient. When an intercept is requested, solve the appropriate axis equation instead of copying the vertex. Look for consistency through symmetry.
Case 1. Vertex of y = (x − 3)² + 2?
(3,2), a minimum.
The squared bracket reaches zero at x = 3.
Case 2. Axis of symmetry of y = (x − 3)² + 2?
The vertical line x = 3.
Points at x = 2 and x = 4 have equal heights.
Case 3. y-intercept of y = (x − 3)² + 2?
(0,11).
Set x = 0, not x = 3.
Case 4. x-intercepts of y = (x − 3)² + 2?
None over the real numbers.
The squared term plus two is always positive.
Case 5. Vertex of y = (x + 3)² + 2?
(−3,2), a minimum.
The bracket vanishes at x = −3.
Case 6. Vertex of y = (x − 4)² − 7?
(4,−7), a minimum.
The inside and outside constants control different directions.
Case 7. Vertex of y = −(x − 4)² + 7?
(4,7), a maximum.
The negative coefficient outside the square reverses orientation.
Case 8. Vertex of y = (x − 1/2)² + 4?
(1/2,4), a minimum.
The same bracket-zero method works with fractional p.
Case 9. Find zeros of y = (x − 3)² − 4.
x = 1 and x = 5.
Both are two units from the symmetry axis x = 3.
Case 10. Find zeros of y = (x − 3)² − 9.
x = 0 and x = 6.
The roots are three units either side of x = 3.
Case 11. Find zero of y = (x − 3)².
x = 3 only.
The parabola touches the x-axis at its vertex.
Case 12. Vertex of y = x² − 6x + 11?
(3,2).
Complete the square as (x − 3)² + 2.
Case 13. Factorise y = (x − 3)² − 4.
(x − 1)(x − 5).
Difference of squares exposes roots one and five.
Case 14. Reflect y = (x − 3)² + 2 to y = −(x − 3)² + 2.
The vertex remains (3,2), but the minimum becomes a maximum.
The external negative changes orientation, not the axis.
Case 15. At x = 2 and 4 on y = (x − 3)² + 2?
Both have y = 3.
The points lie one unit from the axis on opposite sides.
Case 16. At x = 1 and 5 on y = (x − 3)² + 2?
Both have y = 6.
Equal squared horizontal displacements produce equal outputs.
Case 17. For y = 2(x − 3)² + 2, vertex and output at x = 2?
Vertex (3,2); y = 4 at x = 2.
The coefficient changes vertical stretch but not vertex location.
Case 18. For y = −(x + 2)² + 5, vertex?
(−2,5), a maximum.
The bracket is zero at −2; the coefficient is negative.
Case 19. For y = −(x + 2)² + 5, y-intercept?
(0,1).
Set x = 0: −4 + 5 = 1.
Case 20. Does a table at integer x always contain a quadratic minimum?
No. A vertex may occur at a non-integer x such as 2.5.
The algebra identifies a continuous optimum between sampled inputs.
Case 21. What does the vertex (3,2) represent in a restricted model with 4 ≤ x ≤ 6?
The algebraic vertex lies outside the model’s allowed interval.
The interval optimum must be selected from values inside the stated domain.
The clinic intentionally asks similar-looking questions about different mathematical objects. A minimum value is a number, its location is an x-coordinate, the turning point is an ordered pair, and an axis is a line. A student who differentiates these can approach unfamiliar graph questions much more reliably than one who simply memorises a sign convention.
Why does (x − 3)² give a turning point at positive three?
The square is smallest when its bracket is zero. Solving x − 3 = 0 gives x = 3. The printed minus sign is not the vertex coordinate.
What is the turning point of (x − 3)² + 2?
It is the coordinate (3,2). The minimum value is two, achieved when x is three.
Does plus two mean the y-intercept is two?
No. It is the turning point’s y-coordinate. The y-intercept is found at x = 0 and equals eleven for this example.
What happens if there is a minus sign before the square?
The parabola opens downward and the turning point becomes a maximum, provided the outside squared coefficient is negative.
Can a quadratic have a turning point but no x-intercepts?
Yes. The graph y = (x − 3)² + 2 has turning point (3,2), but its smallest y-value is two, so it never meets y = 0.
Is vertex form part of G3 Mathematics for the 2027 SEC?
Yes. The K310 G3 Mathematics syllabus lists quadratic graph properties and sketches involving y = (x − p)² + q and y = −(x − p)² + q. Always check the full official syllabus and teacher’s scope.
Refer to SEAB’s 2027 G3 school-candidate syllabus directory for the authoritative examination outline. Continue through the G1/G2/G3 Mathematics explainer hub and the Clementi Secondary 1 Mathematics tutorial reference for linked foundations and learning methods.