Parabola: vertex, intercepts and distance from a point

Text

Given the parabola $y=x^2-4x-1$:

• (a) find its vertex;
• (b) the intercepts with the $x$- and $y$-axes;
• (c) sketch it;
• (d) find the distance between the vertex and the point $A(3;3)$;
• (e) decide whether $B(1;1)$ belongs to the parabola.

Solution

Coefficients $a=1$, $b=-4$, $c=-1$; $a>0$ opens upward.

(a) Vertex. $x_V=-\frac{b}{2a}=-\frac{-4}{2}=2, \qquad y_V=2^2-4(2)-1=4-8-1=-5$ so $V(2;-5)$.

(b) Intercepts with the axes. $x$-axis: $x^2-4x-1=0 \Rightarrow x=\dfrac{4\pm\sqrt{16+4}}{2}=2\pm\sqrt5\approx 4.236,\ -0.236$. $y$-axis: $x=0 \Rightarrow y=-1$, i.e. $(0;-1)$.

(c) Graph. Parabola opening upward, vertex $V(2;-5)$, crossing the $x$-axis at $2\pm\sqrt5$ and the $y$-axis at $(0;-1)$.

(d) Distance $V$–$A$. With $V(2;-5)$ and $A(3;3)$: $d(V,A)=\sqrt{(3-2)^2+(3+5)^2}=\sqrt{1+64}=\sqrt{65}\approx 8.062$

(e) Does $B(1;1)$ belong? $y(1)=1-4-1=-4\neq1$, so $B$ does not belong to the parabola. $\boxed{V(2;-5)\quad (0;-1)\quad d(V,A)=\sqrt{65}\approx8.062\quad B\notin\text{parabola}}$