How many times does the graph of $x = t^2 - t - 6$, $y = 2t, -5 < t < 5$ cross the $y$-axis?

Question

Although I know that it will pass through $y$-axis twice, when $t = - 2$ at $(0, - 4)$ and when $t = 3$ at $(0, 6)$ but what is the explanation to it?

Answer 1

The graph crossing the $y$-axis for some $t_0$ means that $t_0$ verifies $x(t_0)=t_0^2-t_0-6=0$ (in other worths, when the $x$ coordinate is $0$). So we solve the quadratic equation: $$t^2-t-6=0$$ $$t=\frac{1\pm\sqrt{25}}{2} \longrightarrow t=3, t=-2.$$ This two values for $t$ verify that $-5<t<$, so we conclude your graph crosses the $y$-axis at two diferent points.

We know this two points are $(0,6)$ and $(0,-4)$ because $y=2t$ by your graph's definition, so: $$t=3 \longrightarrow y=6 \longrightarrow \text{point } (0,6)$$ $$t=-2 \longrightarrow y=-4 \longrightarrow \text{point } (0,-4)$$

Answer 2

Eliminating $t$, you see this is an arc of the parabola with equation $$4x=(y-1)^2-25,\quad -10< y<10$$ The intersections of a conic with any line are atmost two. In the present case, the intersections with the $y$-axis are obtained solving $\:(y-1)^2-25=0$.