# How do you graph  f(x) = |x + 3| - 1?

Aug 22, 2017

Vertex $\to \left(x , y\right) = \left(- 3 , - 1\right)$

${y}_{\text{intercept}} = 2$

${x}_{\text{intercept}} \to x = - 4 \mathmr{and} x = - 2$

#### Explanation:

$\textcolor{b l u e}{\text{Explaining what each part of the equation does.}}$
$\textcolor{b l u e}{\text{ie the transformation from basic form}}$

This is transforming the graph of $y = x$ which is the general shape of the sloping line type /

Using the absolute format $y = | x |$ changes the general shape to V where the vertex ( point ) is at the x-axis. So for $\underline{\text{this case}}$:
Vertex $\to \left(x , y\right) = \left(0 , 0\right)$ and y is always positive

Including the + 3 $\to y = | x + 3 |$ 'slides' the graph to the left by 3.
Adding 'slides' to the left whilst subtracting 'slides' it to the right.

Why is this? Consider the graph of $y = | x |$. Suppose we had the x-value of 0. Move to the right by 3. Draw a faint line upwards and note the point on the graph. Now plot a point at that y-value where$x = 0$. You have 'shifted' to the left by 3.

The constant of -1 drops the whole thing by 1.

So for $y = | x + 3 | - 1$ the value if $y$ can now be negative but no less than -1.
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$\textcolor{b l u e}{\text{Determine the vertex}}$

Using the above explanation the Vertex $\to \left(x , y\right) = \left(- 3 , - 1\right)$

Or you cam calculate it by setting $y = - 1$

$- 1 = x + 3 - 1$

$x = - 1 - 3 + 1 = - 3$

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$\textcolor{b l u e}{\text{Determine y-intercept}}$
Set $x = 0$

${y}_{\text{intercept}} = | 0 + 3 | - 1 = 2$

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$\textcolor{b l u e}{\text{Determine x-intercept}}$
Set $y = 0$

$0 = | x + 3 | - 1$

Consider just one side of the V

Set $y = x + 3 - 1 \text{ "->" } y = 0 = x + 2$

In this case $x = - 2$

Compare to the vertex $\to \left(x , y\right) = \left(- 3 , - 1\right)$

The point $x = - 2$ is to the right of $x = - 3$ by 1.

So the other point must be to the left of $x = - 3$ by 1 as well

Thus ${x}_{\text{intercept}} \to x = - 4 \mathmr{and} x = - 2$