A line has a slope of $m,$ and its $y$-intercept is $(0,b)$. If $m = 2$ and $b = 2,$ then what is the $x$-intercept of the line?
Since the $y$-intercept is $(0,b) = (0,2),$ this means that the line passes through the point $(0,2).$ Thus, the $x$-intercept must be $\boxed{1}.$
[asy]
unitsize(1 cm);
pair A, B, C;
A = (0,2);
B = (1,0);
C = (2,0);
draw((-0.5,0)--(2,0));
draw((0,-0.5)--(0,3));
draw(A--(B + C)/2);
label("$y = 2x + 2$", (2,3), NE);
dot("$(0,2)$", A, NW);
dot("$(1,0)$", B, SE);
[/asy]
To find the $x$-intercept of the line, we need to find the point where the line intersects the $x$-axis.
The $x$-intercept is the point at which the $y$-coordinate is 0. So we need to find the value of $x$ when $y = 0$.
We can use the equation of a line, $y = mx + b$, to solve for $x$ when $y = 0$.
Substituting $m = 2$ and $b = 2$ into the equation, we have:
$0 = 2x + 2$
To solve for $x$, we can subtract $2$ from both sides:
$2x = -2$
Finally, we divide both sides by $2$:
$x = -1$
Therefore, the $x$-intercept of the line with slope $m = 2$ and $y$-intercept $(0,2)$ is $(-1, 0)$.
The $x$-intercept of a line occurs when the $y$-coordinate is equal to zero.
To find the $x$-intercept, we can set $y$ equal to zero in the equation of the line.
Since the $y$-intercept is $(0,b),$ we know that the equation of the line is of the form $y = mx + b.$
Substituting $m = 2$ and $b = 2,$ we get $y = 2x + 2.$
Setting $y$ equal to zero, we have $0 = 2x + 2.$
To find $x,$ we can solve this equation for $x.$
Subtracting 2 from both sides, we have $-2 = 2x.$
Dividing both sides by 2, we get $x = -1.$
Therefore, the $x$-intercept of the line is $x = -1.$