How Do You Find the Y-Intercept, X-Intercepts, and Axis of Symmetry of a Parabola?

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These results apply to any quadratic with $a\neq 0$. The coordinates should be read directly from the relevant equation.

The Reasoning

Each feature follows from its definition:

The form of the quadratic can make one feature easier to identify. In factorized form, $y=a(x-p)(x-q)$, the x-intercepts are $(p,0)$ and $(q,0)$. In vertex form, $y=a(x-h)^2+k$, the axis is $x=h$. Converting between general, factorized, and vertex forms lets you check the same features using different representations.

Consider

For the y-intercept, set $x=0$:

Therefore, the y-intercept is $(0,6)$.

For the x-intercepts, set $y=0$:

Thus, $x=1$ or $x=3$, giving $(1,0)$ and $(3,0)$.

For the axis of symmetry,

This equals the midpoint of the roots: $\frac{1+3}{2}=2$. Completing the square also gives $y=2(x-2)^2-2$, whose vertex is $(2,-2)$. This provides a useful check and links the roots to the vertex.

In general, the discriminant $\Delta=b^2-4ac$ determines the number of x-intercepts: two if $\Delta>0$, one repeated intercept if $\Delta=0$, and none if $\Delta<0$.

A common misconception is to give the axis as $(2,0)$. It is a line, so write $x=2$.

IB Exam Technique

This is IB Mathematics AA SL 2.6: Quadratic function, presented in Paper 1 style. Show substitutions and factorization. On Paper 2, a GDC may verify the graph, but the required algebraic setup should still be recorded. If roots do not factorize easily, use the quadratic formula and keep exact values on Paper 1. State intercepts as coordinates, not only root values.