How do you find the slope and y intercept of $y + 1 = 1 (x + 2)$ $y + 1 = 1(x + 2)$ ?

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Answer 1 · 2018-08-02T01:05:35

The slope is 1.

The y-intercept is at (0, 1).

$y + 1 = 1 (x + 2)$ $y + 1 = 1(x+2)$

This equation is in point-slope form:
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Based on the picture, we know that the slope is the value multiplying $x - x_{1}$ $x-x_1$ , so the slope is $1$ $1$ .

To find the $y$ $y$ -intercept, plug in $0$ $0$ for $x$ $x$ and solve for $y$ $y$ :
$y + 1 = 1 (x + 2)$ $y + 1 = 1(x+2)$

$y + 1 = 1 (0 + 2)$ $y + 1 = 1(0+2)$

$y + 1 = 2$ $y + 1 = 2$

$y = 1$ $y = 1$

Therefore, the $y$ $y$ -intercept is at $(0, 1)$ $(0, 1)$ .

Hope this helps!

Answer 2 · 2018-08-02T01:24:04

Slope $1$ $1$ , $y$ $y$ -intercept of $1$ $1$

It may be more intuitive by converting this equation into slope-intercept form

$y = m x + b$ $y=mx+b$ , with slope $m$ $m$ and a $y$ $y$ -intercept of $b$ $b$ . Let's start by distributing the $1$ $1$ on the right to get

$y + 1 = x + 2$ $y+1=x+2$

Next, let's subtract $1$ $1$ from both sides to get

$y = x + 1$ $y=x+1$

We see that our slope is $1$ $1$ , and so is our $y$ $y$ -intercept.

Hope this helps!

How do you find the slope and y intercept of y+1=1(x+2)y + 1 = 1(x + 2)?