Maths › Algebra and functions › Quadratic functions
Quadratic functions
One curve, three costumes: factorised for the roots, completed-square for the vertex, expanded for the y-intercept. And underneath them all sits a single number, the discriminant, that knows how many times the parabola meets the axis before you draw a thing.
Builds on Indices and surds.
IN THIS TOPIC
- Read a quadratic's graph from its factorised, completed-square and expanded forms.
- Use the discriminant to count real roots, and solve for unknowns that control it.
- Solve quadratic equations, including quadratics in a function of the unknown.
WHAT YOU PROBABLY THINK
If b² − 4ac comes out negative, you made a mistake.
Three forms, one parabola
A quadratic function f(x) = ax2 + bx + c draws a parabola, opening upward when a > 0 and downward when a < 0, and each written form answers a different question at sight. The expanded form shows the y-intercept, c. The factorised form shows the roots. And completing the square shows the vertex, because
puts all the x-dependence inside one squared bracket. A square is smallest when its bracket is zero, so the vertex sits at x = −b/2a, and whatever constant is left over is the minimum, or maximum, value itself.
WORKED EXAMPLE
Completing the square with a leading coefficient
Express 2x2 − 12x + 7 in the form a(x + p)2 + q, and state the minimum point of the curve.
Take the 2 out of the x terms: 2(x2 − 6x) + 7 = 2[(x − 3)2 − 9] + 7.
Distribute: 2(x − 3)2 − 18 + 7 = 2(x − 3)2 − 11.
The bracket is zero at x = 3, so the minimum point is (3, −11). Expanding back is a ten-second check, and worth every one of the ten seconds.
The discriminant
Feed a quadratic to the formula, which is on the must-learn list,
and everything interesting happens under the root. The quantity b2 − 4ac is the discriminant, and its sign is a complete census of the roots: positive means two distinct real roots, zero means one repeated root, negative means no real roots. No real roots is a perfectly healthy answer, not an error, which retires the lie above: it simply means the parabola never reaches the axis.
WORKED EXAMPLE
Choosing k for a repeated root
Find the values of k for which x2 + kx + 9 = 0 has a repeated root.
Repeated root means discriminant zero: k2 − 4 × 1 × 9 = 0, so k2 = 36.
k = 6 or −6: two different parabolas, each just touching the axis.
Questions that give a quadratic a parameter are almost always discriminant questions underneath; translate the root condition into a sign condition first.
YOUR TURN
A condition, not a value
Find the set of values of k for which x2 + 3x + k = 0 has real roots, before opening the working.
Show the working
Real roots means b2 − 4ac ≥ 0: here 9 − 4k ≥ 0.
So k ≤ 9/4, or in set notation {k : k ≤ 9/4}. The ≥ matters: “real roots” includes the repeated case, so the boundary value stays in.
Quadratics in disguise
The examiners' favourite twist is a quadratic in a function of the unknown: an equation that becomes ordinary once the right substitution is named. Powers, roots, and later trigonometric and exponential expressions can all play the role of the hidden variable; last lesson's x − 6√x + 4 = 0 was one of them.
WORKED EXAMPLE
An exponential with a quadratic inside
Solve 4x − 7(2x) + 12 = 0.
Since 4x = (2x)2 by the index laws, substitute u = 2x: u2 − 7u + 12 = 0.
Factorise: (u − 3)(u − 4) = 0, so u = 3 or u = 4.
Return to x: 2x = 3 gives x = log2 3, and 2x = 4 gives x = 2. One exact logarithm, one integer, both valid since each u is positive.
The rhythm never changes. Spot the hidden square, name u, solve the visible quadratic, translate back, and check each u makes sense for the substitution.
TRY IT UNSEEN
A quartic that is secretly a quadratic
Solve x4 − 5x2 + 4 = 0.
Show the working
Substitute u = x2: u2 − 5u + 4 = (u − 1)(u − 4) = 0, so u = 1 or u = 4.
Translate back: x2 = 1 gives x = ±1; x2 = 4 gives x = ±2. Four roots: x = −2, −1, 1, 2.
The translation step doubled the answers, because each positive u has two square roots. The substitution is where solutions are gained or lost, so it deserves the care.
THE EXAM BIT
- Completing the square with a ≠ 1: take the coefficient out of the x terms only, complete inside, then distribute back. Expanding to check costs seconds and saves marks.
- State the discriminant condition before computing it: “real and distinct means b2 − 4ac > 0”. The inequality chosen is itself a mark.
- “Real roots” includes the repeated case (≥); “distinct real roots” excludes it (>). Read the wording twice.
- For hidden quadratics, write the substitution down explicitly, u = 2x or u = x2, and finish by translating every u back to x, discarding impossible values with a reason.
- Choose the solving tool to fit: factorise when it is quick, complete the square when the vertex or exact surds are wanted, formula when nothing factorises.
CHECK YOURSELF
Express 3x2 + 6x − 2 in the form a(x + p)2 + q, state the minimum point, and use the discriminant to say how many real roots 3x2 + 6x − 2 = 0 has.
Show a hint
Take out the 3 first; the discriminant only needs a, b and c.
Show the answer
3(x2 + 2x) − 2 = 3[(x + 1)2 − 1] − 2 = 3(x + 1)2 − 5.
The minimum point is (−1, −5): bracket zero at x = −1, leftover constant −5.
b2 − 4ac = 36 + 24 = 60 > 0, so the equation has two distinct real roots, which agrees with a minimum below the axis on an upward parabola.
Completed square names the vertex; the leftover constant is the extreme value.
The discriminant counts the roots before any solving: > 0 two, = 0 one, < 0 none.
CHECK YOUR PROGRESS
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- Read a quadratic's graph from its factorised, completed-square and expanded forms.
- Use the discriminant to count real roots, and solve for unknowns that control it.
- Solve quadratic equations, including quadratics in a function of the unknown.
No animated video for this topic yet; these notes stand alone.