6.40 Given a quadratic function $y=f(x)$ whose graph has a vertex at $(-1,1)$ and intersects the $y$-axis at $(0,2)$.
(1) Find the expression for this quadratic function;
(2) When $x=8, y=$ ?
(3) For any given value of $y$, can we always find a value of $x$? Why? Explain using a graph.
[Solution] (1) Since $(-1,1)$ is the vertex of $y=f(x)$, we can assume $y=a(x+1)^{2}+1$. Given that the intersection with the $y$-axis is $(0,2)$, we have
$$2=a(0+1)^{2}+1 \text {, solving for } a=1 \text {. }$$

Therefore, the quadratic function is
$$y=(x+1)^{2}+1$$

which is $y=x^{2}+2 x+2$.
(2) When $x=8$, $y=9^{2}+1=82$.
(3) For any given value of $y$, it is not always possible to find a corresponding value of $x$. Since $y=(x+1)^{2}+1 \geqslant 1$, when $y$ is less than 1, there is no corresponding $x$ value.

As shown in the figure, the line $y=m$, when $m<1$, does not intersect the parabola $y=x^{2}+2 x+2$.