Quadratic GraphsCambridge IGCSE Maths: Revision notes

A quadratic graph is the graph of a function of the form $y=ax^2+bx+c$ (or $y=\pm x^2+ax+b$). Every quadratic graph is a parabola — a symmetric curve shaped like a U (if $a>0$) or an upside-down U, ∩ (if $a<0$).

Since these questions must be solved without a picture, you should be able to construct a table of values by substituting $x$-values into the equation, and describe/interpret the graph's features purely from the algebra.

Every parabola has a line of symmetry — a vertical line through its turning point, splitting the curve into two mirror-image halves.

For $y=ax^2+bx+c$, the line of symmetry (and the $x$-coordinate of the turning point) is: $x = -\frac{b}{2a}$

Once you have this $x$-value, substitute it back into the equation to find the $y$-coordinate of the turning point.

Example: For $y=x^2-4x+3$: line of symmetry $x=-\frac{-4}{2(1)}=2$. Substituting: $y=4-8+3=-1$. Turning point is $(2,-1)$.

Completing the square rewrites $y=ax^2+bx+c$ in the form $y=a(x+p)^2+q$, from which the turning point can be read directly as $(-p, q)$.

1. Factor out $a$ from the $x^2$ and $x$ terms if $a \neq 1$
2. Halve the coefficient of $x$ (inside the bracket) to find $p$
3. Add and subtract the correction term to keep the expression equivalent

Example: Write $y=x^2-6x+11$ in completed-square form. $x^2-6x = (x-3)^2-9$, so $y=(x-3)^2-9+11=(x-3)^2+2$. Turning point: $(3, 2)$, a minimum since $a=1>0$.

The roots of a quadratic (where $y=0$) can be found by factorising, completing the square, or using the quadratic formula.

Solving equations "graphically" (finding intersections of graphs) is interpreted algebraically here: to find where two graphs meet, set their equations equal to each other and solve.

Example: Find where $y=x^2-1$ and $y=2x+2$ intersect. $x^2-1=2x+2 \Rightarrow x^2-2x-3=0 \Rightarrow (x-3)(x+1)=0 \Rightarrow x=3$ or $x=-1$. Substituting gives intersection points $(3,8)$ and $(-1,0)$.