Recognising a parabola and a line from the graph and their intersections

Text

The graph shows a parabola (with vertex at $(-2;0)$, passing through $(0;-2)$) and a line. Answer:

• (a) prove that the drawn parabola has equation $y=-\dfrac{x^2}{2}-2x-2$;
• (b) find the equation of the drawn line, passing through $(0;-4)$ and $(-2;0)$;
• (c) find algebraically the intersections between line and parabola.

Solution

(a) From the vertex $V(-2;0)$ write the parabola as $y=a(x+2)^2$. Impose passage through $(0;-2)$: $-2=4a\Rightarrow a=-\dfrac12$. Expanding: $y=-\frac12(x+2)^2=-\frac{x^2}{2}-2x-2.$ (b) The line through $(0;-4)$ and $(-2;0)$ has slope $m=-2$ and intercept $-4$: $y=-2x-4.$ (c) Intersections: $-\dfrac{x^2}{2}-2x-2=-2x-4$, i.e. $-\dfrac{x^2}{2}+2=0\Rightarrow x=\pm2$. For $x=-2$: $y=0$, point $(-2;0)$; for $x=2$: $y=-8$, point $(2;-8)$. $\boxed{(-2;0)\ \text{and}\ (2;-8).}$