# What is the slope and intercept for x - y + 1 = 0 and how would you graph it?

Oct 3, 2015

Slope: $1$
y-intercept: $1$
x-intercept: $\left(- 1\right)$

#### Explanation:

The general slope-intercept form for a line is
$\textcolor{w h i t e}{\text{XXX}} y = m x + b$
$\textcolor{w h i t e}{\text{XXXXX}}$where $m$ is the slope and $b$ is the y-intercept

$x - y + 1 = 0$
can be converted into slope-intercept form by
adding $y$ to both sides and then exchanging the sides:
$\textcolor{w h i t e}{\text{XXX}} x + 1 = y$
$\textcolor{w h i t e}{\text{XXX}} y = \left(1\right) x + 1$
$\textcolor{w h i t e}{\text{XXXXX}}$Note that I have inserted the implied coefficient of $1$ for $x$

Based on the general form we can see that
$\textcolor{w h i t e}{\text{XXX}}$the slope is $m = 1$
and
$\textcolor{w h i t e}{\text{XXX}}$the y-intercept is $b = 1$

Assuming the x-intercept is also required,
we note that the x-intercept is the value of $x$ when $y = 0$
$\textcolor{w h i t e}{\text{XXX")x-(0)+1=0color(white)("XX")rarrcolor(white)("XX}} x = - 1$

The x and y-intercepts give us the points
$\textcolor{w h i t e}{\text{XXX}} \left(- 1 , 0\right)$ and $\left(0 , 1\right)$ respectively.
If we plot these two points on the Cartesian plane and draw a straight line through them, we will obtain the required graph

graph{(x-y+1)((sqrt(x^2+(y-1)^2))-0.1)((sqrt((x+1)^2+y^2))-0.1)=0 [-5.25, 5.85, -2.02, 3.527]}