Functions question help me? a) We have the function $f(x,y)=x-y+1.$ Find the values of $f(x,y)$ in the points of the parabola $y=x^2$ and build the graph $F(x)=f(x,x^2)$ .
So, some points of the parabola are $(0;0), (1;1), (2;4)$. I replace these in $f(x,y)$ and I have $f(x,y)=1,1,-1\dots$. The graph $f(x,x^2)$ must have the points $(0;1),(1;1)$ and $(2;-1)\,$ , right?
b)We have the function $f(x,y) =\sqrt x + g[\sqrt x-1]$.Find $g(x)$ and $f(x,y)$ if $f(x,1)=x$? 
I dont even know where to start here :/
 A: Hints:
a) $F(x)=f(x,x^2)=x-x^2+1 =-(x-\frac12)^2+...$
b) As it stands, if I understand well, $f(x,y)=\sqrt x+g(\sqrt x-1)$ is independent from $y$. Then $f(x,1)=x$ means
$$\sqrt x+g(\sqrt x-1)=x\,.$$
To find $g$, probably the best is to introduce a new variable $z:=\sqrt x-1$, then $\sqrt x=z+1$ and $x=(z+1)^2$.
A: And If  the function $f(x,y)$ be like $$f(x,y)=\sqrt{y}+g(\sqrt{x}-1)$$, then the way is similar to @Berci's answer.Y ou pointed that $f(x,1)=x$ so $$1+g(\sqrt{x}-1)=x$$ Set $\sqrt{x}-1=t$ then $x=(t+1)^2$ and so $$g(t)=(t+1)^2-1=t^2+2t$$ Now you can easily find the function $f(x,y)$.
A: (a) 


*

*To find points on $f(x, y)$ that are also on the parabola $y = f(x) = x^2$: Solve for where $f(x, y) = x - y + 1$ and $y = f(x) = x^2$ intersect by putting $f(x, y) = f(x)$:
$$ x^2 = x - y + 1$$ and and express as a value of $y$:


$$y = 1 + x - x^2\;\;\text{ and note}\;\; y = F(x) = f(x, x^2)\tag{1}$$


*

*Note that this function $F(x)$ is precisely $f(x, x^2)$. Find enough points in this equation to plot it. (The points that satisfy $(1)$ will be points on $f(x,y)$ which satisfy $y = x^2$.)  You will find that $\;\;F(x) = 1 + x - x^2\;\;$ is also a parabola. (See image below.) We can manipulate $F(x)$ to learn where the vertex of the parabola is located: write $F(x): (y - \frac 54) = -(x - \frac 12)^2$, so the vertex is located at $(\frac 12, \frac 54)$. The negative sign indicates that the parabola opens downward.


$\quad F(x) = 1 + x - x^2:$


