How do you graph and find the vertex for $y = 2 | x - 4 | + 1$ $y=2abs(x-4)+ 1$ ?

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Answer 1 · 2015-07-11T21:13:47

The vertex is at ( $4, 1$ $4,1$ ).

The standard form for an absolute value equation is

$y = a | x - h | + k$ $y = a|x-h| + k$

Your equation is

$y = 2 | x - 4 | + 1$ $y = 2|x−4| + 1$

So

$a = 2$ $a = 2$ , $h = 4$ $h = 4$ , and $k = 1$ $k =1$

Vertex

The vertex is at $x = –h = 4$ .

The $y$ -coordinate of the vertex is at $y = k = 1$ .

The vertex is at ( $4,1$ ).

Graph

Now we prepare a table of $x$ and $y$ values.

The axis of symmetry passes through $x = 4$ .

Let's prepare a table with points that are 5 units on either side of the axis, that is, from $x = -1$ to $x = 9$ .

Table Table

Plot these points.

graph{2|x-4|+1 [-1, 10, -1, 10]}

And we have our graph.

How do you graph and find the vertex for y=2abs(x-4)+ 1y=2|x−4|+1 y=2abs(x-4)+ 1?