Quadratic functions and graphsAQA A-Level Maths: Revision notes

A quadratic function has the form $f(x)=ax^2+bx+c$ with $a\neq0$. Its graph is a parabola: U-shaped (a minimum) when $a>0$, and an inverted U (a maximum) when $a<0$. Key features to find:

• the $y$-intercept: $(0,c)$;
• the roots (where the graph meets the $x$-axis): solve $ax^2+bx+c=0$;
• the line of symmetry, $x=-\frac{b}{2a}$, halfway between the roots;
• the turning point (vertex), on the line of symmetry.

Completing the square rewrites $ax^2+bx+c$ in the form $a(x+p)^2+q$. When $a=1$: $x^2+bx+c=\left(x+\frac b2\right)^2-\frac{b^2}{4}+c$. For example $x^2-6x+13=(x-3)^2-9+13=(x-3)^2+4$. When $a\neq1$: take out the factor $a$ from the $x^2$ and $x$ terms first. $2x^2+8x-3=2\left(x^2+4x\right)-3=2\left[(x+2)^2-4\right]-3=2(x+2)^2-11$. From $a(x+p)^2+q$ the turning point is $(-p,\,q)$ and the line of symmetry is $x=-p$. It is a minimum if $a>0$ and a maximum if $a<0$.

For $ax^2+bx+c=0$ the discriminant is $\Delta=b^2-4ac$, the part under the root in the quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

• $b^2-4ac>0$: two distinct real roots (the graph crosses the $x$-axis twice).
• $b^2-4ac=0$: one repeated root (the graph touches the $x$-axis at its turning point).
• $b^2-4ac<0$: no real roots (the graph does not meet the $x$-axis). For $x^2-6x+13$: $\Delta=36-52=-16<0$, so there are no real roots. The minimum value is $4>0$, which agrees.

Often a question gives a quadratic containing a constant $k$ and asks for conditions on $k$. Worked example. For $x^2-kx+(k+3)=0$, $\Delta=k^2-4(k+3)=k^2-4k-12=(k-6)(k+2)$.

• Repeated root: $\Delta=0$, so $k=6$ or $k=-2$.
• Two distinct real roots: $\Delta>0$, so $(k-6)(k+2)>0$. Sketch $y=(k-6)(k+2)$: it is above the axis outside the roots, so $k<-2$ or $k>6$.
• No real roots: $\Delta<0$, so $-2<k<6$.

Maximum and minimum problems often lead to a quadratic. Write the quantity in terms of one variable, complete the square, and read off the turning point. Worked example. 40 m of fencing encloses a pen against a wall. With the perpendicular sides $x$, the area is $A=x(40-2x)=40x-2x^2=200-2(x-10)^2$. The maximum area is $200$ m$^2$ at $x=10$. Remember the domain: the lengths must be positive, so $0<x<20$. To find when the area is at least 150, solve $40x-2x^2\geq150$ to get $5\leq x\leq15$. The discriminant shows a target is impossible: $A=210$ gives $\Delta<0$.