Quadratic functions and their graphsIB MYP Maths Standard: Revision notes

A quadratic function has the form $y=ax^2+bx+c$ with $a\ne0$. Its graph is a smooth curve called a parabola. If $a>0$ the parabola is U-shaped with a lowest point (minimum). If $a<0$ it is $\cap$-shaped with a highest point (maximum). The parabola is symmetrical about a vertical line, the axis of symmetry.

Choose values of $x$, work out $y$, plot the points and join them with a smooth curve (never straight segments). For $y=x^2-2x-3$: using $x=-2,-1,0,1,2,3,4$ gives $y=5,0,-3,-4,-3,0,5$. The values repeat on either side of $x=1$, which shows the symmetry. A graphing calculator or graphing software draws the same curve and lets you read off key points.

The roots (or $x$-intercepts) are the values of $x$ where the curve meets the $x$-axis, so $y=0$. Solve $ax^2+bx+c=0$, for example by factorising: $x^2-2x-3=(x-3)(x+1)=0$ gives $x=3$ or $x=-1$. A parabola may have two roots, one root (touching the axis) or none. The $y$-intercept is where the curve crosses the $y$-axis, so $x=0$: it is the constant $c$.

The axis of symmetry is halfway between the roots: $x=\frac{x_1+x_2}{2}$. (If there are no roots, use the formula $x=-\frac{b}{2a}$.) The vertex is the point of the curve on the axis: substitute that $x$-value into the function. For $y=x^2-2x-3$ the roots are $-1$ and $3$, so the axis is $x=1$ and $y=1-2-3=-4$: the vertex is $(1,-4)$, a minimum. To sketch a parabola, mark the intercepts and the vertex and draw a smooth curve through them.

Many real situations follow a parabola: the path of a ball, the area of a field, profit. Read the key points in context. For $h=20t-5t^2$, the roots $t=0$ and $t=4$ are when the ball is kicked and when it lands. The vertex $(2,20)$ means the ball reaches its maximum height of $20$ m after $2$ s. Always give units, and check which values are sensible: a negative length or time is not possible, so the graph only applies for a limited range of $x$ (the domain).