Finding the value of the y-intercept of a line and a function The question is actually in Dutch, and so I hope I will translate it well. The question is:
For what value of b does the line $y = x + b$ "hit" the graph of the function $f(x) = x^2-2x+3$?
My assumption here is that I'm dealing with a question that will involve calculus. I can find $f'(x) = 2x - 2$, but I'm lost with what to do next, as there are no points to go on. I do have a slope of 1, but then I also don't know what to do with that.
So, in short, I need some help in getting to the next step!
 A: I assume the "hit" means it touches the quadratic function at one point.
That means, at that point, they have the same slope. So you need to find at what point the derivative (slope of the tangent line) of the quadratic function is $1$. Once you find that, you can easily find $b$.
A: $$
f(x) = x^2-2x+3\\
g(x) = x + b
$$
when will these two hit, is when 
$$
f(x) = g(x)\\
x^2-2x+3 = x + b \implies x^2 - 3x + 3-b = 0
$$
what values of $b$ does this occur?
A: A non-calculus solution:
They intersect when:
$x^2-2x+3=x+b$
Or when 
$x^2-3x+3-b=0$
This in general has two solutions in $x$ (sometimes 0 real solutions); it has one solution only when we can factor this as a perfect square: $(x-k)^2$.
The coefficient on $x$, 3, implies that $k=\frac{3}{2}$.
Thus, $x+b$ intersects the parabola exactly once if and only if
$x^2-3x+3-b=(x-\frac{3}{2})^2$,
which, simplifying, means
$3-b=\frac{9}{4}\Leftrightarrow b=\frac{3}{4}$.
A: tangent line of $f(x)$ at a point $(a,f(a))$ is  $y=f(a)+f'(a)(x-a)$.
Rearranging we have  $y=f'(a) x +f(a)-f'(a)a $
Comparing with your line we must have $f'(a)=2a-2=1$. Hence $a=3/2$ and $b=f(3/2)-3/2.$
