The absolute value function

[Studyinghard]

This is just a tad confusing and a big at the moment actually. It has stopped my progression into the course.

y =

Can someone please do step by step for x intercept, y intercepts etc. and a graph would be just the best

[TrueTears]

If then



If then



Sketch those 2 hybrid equations with the domain I gave. There's your graph

[Studyinghard]

Sorry, totally lost. I was more looking for a find x int and y int process.

I have uploaded the answer with paint and am just confused with well the whole graph. When I think I have the right answer it seems a mere fluke that I got it right.

[Aqualim]

If it helps, break it down into easier steps for yourself. so instead of solving the whole thing at once, start with .
When
When
Endpoint;

Now because the graph is an absolute function, we must change the y-points so they are positive,
when
Endpoint;

Now just simply sketch the graph, but remember when the graph hits , the line not continue into the negative numbers

Now to include the part, just add 3 units onto 0.6 giving 3.6 and 3 units onto 2, giving 5

And it should look like the graph you have drawn

[Ilovemathsmeth]

For modulus functions, you always need to evaluate the rule and then sketch. Especially for something twisted like that. First find where the interior function (inside modulus) is greater than zero. The function remains the same for the sections that it's greater than zero. Then find where it is negative. Where it is negative, you must negate the interior function thereby reflecting it into the x-axis and making that part positive. Negate only the interior of the modulus after removing the modulus. My LaTex is shaky, so I can't show you working

[Studyinghard]

If it helps, break it down into easier steps for yourself. so instead of solving the whole thing at once, start with .
When
When
Endpoint;

Now because the graph is an absolute function, we must change the y-points so they are positive,
when
Endpoint;

Now just simply sketch the graph, but remember when the graph hits , the line not continue into the negative numbers

Now to include the part, just add 3 units onto 0.6 giving 3.6 and 3 units onto 2, giving 5

And it should look like the graph you have drawn

Well with that in mind and the same procedure I went about dissecting the next one.



starting with the easy part
When y = 0, x = 63
When x = 0, y = -7
endpoint (-1,-8)

Now I am stuck on the theory a bit, because it is a negative graph should all the y points b negative?

Also in the answer below, how are there x intercepts of 35 and 99?

[the.watchman]

Breaking it down:

Take first:

It is entirely positive, intercepts at (0,7) and (63,0)

Then consider :

It is entirely negative with intercepts now at (0,-7) and (63,0)

Now add 2 to the expression to get :

The line of reflection is now , not

So there are two x-intercepts:

Let



Two cases:

,

OR ,

SO the x-ints are at (35,0) and (99,0)

[Studyinghard]

Breaking it down:

Take first:

It is entirely positive, intercepts at (0,7) and (63,0)

Then consider :

It is entirely negative with intercepts now at (0,-7) and (63,0)

Now add 2 to the expression to get :

The line of reflection is now , not

So there are two x-intercepts:

Let



Two cases:

,

OR ,

SO the x-ints are at (35,0) and (99,0)

ooo nifty. might go and try some more of these. Thanks