1
Question
How do you graph f(x)=|x+3|−1?
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Solution
Vertex →(x,y)=(−3,−1)
yintercept=2
xintercept→=−4 and =−2
Explanation:
Explaining what each part of the equation does. i.e. 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 this case:
Vertex →(x,y)=(0,0) and y is always positive.
Including the +3→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 o 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' tot 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
Determine the vertex
Using the above explanation the vertex →(x,y)=(−3,−1)
Or you can calculate it by setting y=−1
−1=x+3−1
x=−1−3+1=−3
Determine y-intercept
Set x=0
yintercept=|0+3|−1=2
Determine x-intercept
Set y=0
0=|x+y|−1
Consider just one side of the V
Set y=x+3−1→y=0=x+2
In the case x=−2
Compare to the vertex →(x,y)=(−3,−1)
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 xintercept→x=−4 and x=−2
yintercept=2
xintercept→=−4 and =−2
Explanation:
Explaining what each part of the equation does. i.e. 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 this case:
Vertex →(x,y)=(0,0) and y is always positive.
Including the +3→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 o 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' tot 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
Determine the vertex
Using the above explanation the vertex →(x,y)=(−3,−1)
Or you can calculate it by setting y=−1
−1=x+3−1
x=−1−3+1=−3
Determine y-intercept
Set x=0
yintercept=|0+3|−1=2
Determine x-intercept
Set y=0
0=|x+y|−1
Consider just one side of the V
Set y=x+3−1→y=0=x+2
In the case x=−2
Compare to the vertex →(x,y)=(−3,−1)
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 xintercept→x=−4 and x=−2
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