$y= 2x^2+8-1$: why does not touch the $y$ axis? If $y=0$ then  $2x^2+7=0$; $2x^2=-7$; $x^{2}= -\frac{7}{2}.$; if $x=0$ then $x =-\frac{7}{2}$ right?
Please someone explain to me this concept, although it seems a dumb question?
Since $x\geq 0$ for all $x\in \mathbb R$ we have that $y=2x^{2}+7\geq7&gt;0.$
So the minimum value of $y$ is $7$ and this occurs when $x=0$. Indeed when $x=0$ we have $y=2(0)^{2}+7=7$, which is where the quadratic crosses the $y$-axis.
If you set $y=0$, then this gives you the values where the graph crosses the $x$-axis, but $y\geq7$, so it will never cross the $x$-axis. Usually you would compute the discriminant to find the number of roots.
Below is a sketch:
In fact it touches the y axis (x=0). The intersection of the curve and the y axis is at the point (0,7).
The curve does not touch the x axis (y=0) because in this case the resulting polynomial has two imaginary solutions (one is the complex conjugate of the other).