Vertex, roots and intercept of a parabola

Text

Given the parabola $y=x^{2}-4x+3$, find the vertex, the roots and the intersection with the $y$-axis.

Solution

Vertex. The $x$-coordinate is $x_V=-\dfrac{b}{2a}=-\dfrac{-4}{2}=2$; the $y$-coordinate is $y_V=2^{2}-4\cdot 2+3=-1$. So $V(2,-1)$. Roots. Solve $x^{2}-4x+3=0$: $(x-1)(x-3)=0\Rightarrow x=1\ \lor\ x=3$. Intersection with the $y$-axis. Setting $x=0$: $y=3$, i.e. the point $(0,3)$. $\boxed{\,V(2,-1),\ \text{roots } x=1,\,x=3,\ y\text{-axis}:(0,3)\,}$