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Edexcel IGCSE·Maths·Edexcel IGCSE Maths

Quadratic & Other Graphs

5 min read

Parabolas, cubics, reciprocals, exponentials and circles, and reading real-life graphs.

Recognising a Quadratic Graph

A quadratic function has the form y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c where a≠0a \neq 0a=0. Its graph is always a parabola — a smooth, symmetrical U-shape.

    If a>0a > 0a>0 the parabola opens upwards (a valley, with a minimum point).
    If a<0a < 0a<0 it opens downwards (a hill, with a maximum point).

The bigger ∣a∣|a|∣a∣ is, the narrower the curve. The constant ccc is the yyy-intercept, because when x=0x = 0x=0 we get y=cy = cy=c.

Key terms Parabola — the U-shaped curve of any quadratic.

Root — a value of xxx where the curve crosses the xxx-axis, i.e. where y=0y = 0y=0.

Turning point — the single lowest (minimum) or highest (maximum) point.

Line of symmetry — the vertical line through the turning point; the curve is a mirror image across it.

Roots, Turning Point and Symmetry

root root turning pt line of symmetry x y
A parabola showing its two roots, line of symmetry and minimum turning point

To plot a quadratic, build a table of values, substituting each xxx carefully (watch negative-squared signs), then join the points with a smooth curve — never straight line segments.

The Turning Point by Completing the Square

Completing the square rewrites the quadratic as y=a(x−h)2+ky = a(x - h)^2 + ky=a(x−h)2+k. The turning point is then read straight off as (h, k)(h,\,k)(h,k), and the line of symmetry is x=hx = hx=h.

Worked example Find the turning point and line of symmetry of y=x2−6x+11y = x^2 - 6x + 11y=x2−6x+11.

Halve the xxx-coefficient: −6÷2=−3-6 \div 2 = -3−6÷2=−3.

Exam tip The line of symmetry always sits exactly halfway between the two roots. If you know the roots are x=1x = 1x=1 and x=5x = 5x=5, the symmetry line is x=1+52=3x = \tfrac{1+5}{2} = 3x=21+5​=3 — no algebra needed.

Solving Quadratics Graphically

The roots of ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0 are simply where the graph y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c crosses the xxx-axis. Read those xxx-values off the plot.

To solve a different equation, say x2−3x−1=0x^2 - 3x - 1 = 0x2−3x−1=0 using a graph of y=x2−3x−4y = x^2 - 3x - 4y=x2−3x−4, rearrange so one side matches the drawn curve:

x2−3x−1=0  ⇒  x2−3x−4=−3x^2 - 3x - 1 = 0 \;\Rightarrow\; x^2 - 3x - 4 = -3x2−3x−1=0⇒x2−3x−4=−3

So draw the straight line y=−3y = -3y=−3 and read the xxx-coordinates where it meets the curve. Those intersection points are the solutions.

Watch out A quadratic can have two, one, or zero real roots. One root means the curve just touches the axis (the turning point sits on it); zero roots means the whole curve floats above or below the axis.

Cubic Graphs

A cubic has the form y=ax3+bx2+cx+dy = ax^3 + bx^2 + cx + dy=ax3+bx2+cx+d. Its graph has a characteristic stretched-S shape with up to two turning points and can cross the xxx-axis up to three times.

    If a>0a > 0a>0 the curve rises from bottom-left to top-right.
    If a<0a < 0a<0 it falls from top-left to bottom-right.
origin x y y = x³
A positive cubic y = x³ rising left to right through the origin

Reciprocal Graphs

A reciprocal graph has the form y=kxy = \dfrac{k}{x}y=xk​ (with k≠0k \neq 0k=0). It forms two separate curves called a hyperbola.

    The graph never touches the axes — they are asymptotes that the curve approaches but never reaches.
    For k>0k > 0k>0 the branches sit in the top-right and bottom-left.
    There is no value at x=0x = 0x=0, since dividing by zero is undefined.

Exponential Graphs

An exponential graph has the form y=k axy = k\,a^xy=kax where a>0a > 0a>0. This models growth (when a>1a > 1a>1) and decay (when 0<a<10 < a < 10<a<1).

    The curve always passes through (0,k)(0, k)(0,k), because a0=1a^0 = 1a0=1.
    It rises (or falls) ever more steeply and has the xxx-axis as a horizontal asymptote — it gets very close to but never touches y=0y = 0y=0.
(0, 1) x y y = 2ˣ asymptote
Exponential growth y = 2ˣ passing through (0,1) with the x-axis as asymptote

Key terms Asymptote — a line the curve approaches infinitely closely but never meets. Reciprocal and exponential graphs both have them.

The Graph of a Circle

The equation x2+y2=r2x^2 + y^2 = r^2x2+y2=r2 gives a circle centred on the origin with radius rrr. For example x2+y2=25x^2 + y^2 = 25x2+y2=25 is a circle of radius 555, since 25=5225 = 5^225=52.

Exam tip Don't mistake x2+y2=25x^2 + y^2 = 25x2+y2=25 for a quadratic curve. The right-hand side is the radius squared, so the radius is 25=5\sqrt{25} = 525​=5, not 252525.

Gradients and Areas from Graphs

For a curve, the gradient at a point is found by drawing a tangent — a straight line just touching the curve there — and calculating its gradient as riserun\dfrac{\text{rise}}{\text{run}}runrise​ using two clear points on the tangent.

The area under a curve between two xxx-values can be estimated by splitting the region into strips (trapezia or counting squares) and adding them up.

Real-Life Graphs

These ideas have direct physical meaning in motion graphs.

Graph typeGradient meansArea under means
Distance–timespeed(nothing useful)
Speed–timeaccelerationdistance travelled
    On a distance–time graph, a steeper line is faster; a horizontal line is stationary; a curve means changing speed.
    On a speed–time graph, a positive gradient is acceleration, a negative gradient is deceleration, and the area underneath gives total distance.

Worked example A car travels at a steady 20 m/s20\ \text{m/s}20 m/s for 30 s30\ \text{s}30 s, shown as a horizontal line on a speed–time graph. The distance is the area of the rectangle beneath it:

distance=20×30=600 m\text{distance} = 20 \times 30 = 600\ \text{m}distance=20×30=600 m.

If the line then slopes down to 000 over the next 10 s10\ \text{s}10 s, that braking distance is a triangle: 12×10×20=100 m\tfrac{1}{2} \times 10 \times 20 = 100\ \text{m}21​×10×20=100 m, giving 700 m700\ \text{m}700 m in total.

Real world Engineers read speed–time graphs straight off a vehicle's data to find both how hard it accelerated (gradient) and how far it went (area) — the same maths you use in the exam runs inside a car's safety systems.

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y=(x−3)2−9+11=(x−3)2+2y = (x - 3)^2 - 9 + 11 = (x - 3)^2 + 2y=(x−3)2−9+11=(x−3)2+2.

The bracket (x−3)2(x-3)^2(x−3)2 is smallest (zero) when x=3x = 3x=3, giving y=2y = 2y=2.

So the minimum turning point is (3, 2)(3,\,2)(3,2) and the line of symmetry is x=3x = 3x=3. Since the minimum yyy is 2>02 > 02>0, the curve never reaches the xxx-axis, so there are no real roots.