I need help with this graph, pls help!

[fabulousunicorn]

g(u)= {u-1 if u <0; u² -2u - 3 if 0 ≤ u ≤ 3; 0 if u > 3

can you explain why there is ordered pair (0,-1) when u < 0? and why the range is {y|y ≤ 0} but not {y|-4 ≤ y ≤ 0}

 

[pka]

g(u)= {u-1 if u <0; u² -2u - 3 if 0 ≤ u ≤ 3; 0 if u > 3

You need to read posting instructions before posting.
We do not offer to tutor you. So show work & ask questions.
Perhaps if posted correctly you could understand piecewise defined functions?
\(\displaystyle {u(x)=\begin{cases}u-1&\: u<0\\u^2-2u-3 &\:0\le u\le 3\\0 &\: 3 < u \end{cases}}\)

 

[stapel]

\(\displaystyle \mbox{Graph the fun}\mbox{ction: }\, g(u)\, =\, \begin{cases}u\, -\, 1,&\mbox{if }\, u\, <\, 0 \\ u^2\, -\, 2u\, -\, 3,&\mbox{if }\, 0\, \leq\, u\, \leq\, 3 \\ 0,&\mbox{if }\, u\, >\, 3 \end{cases}\)

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There are three equations to contend with, depending on the value of the domain, u. So I will graph three equations on the same axis starting with negative u-values and working up to u-values greater than 3. On the graph, u-values are on the x-axis. Below is a chart of ordered pairs for the equation.

u < 0 g(u) = u - 1 -2 -2 - 1 = -3 -1 -1 - 1 = -2 0 0 - 1 = -1

0 < u < 0 g(u) = u2 - 2u - 3 0 0 - 0 - 3 = -3 1 1 - 2 - 3 = -4 2 4 - 4 - 3 = -3 3 9 - 6 - 3 = 0

u > 3 g(u) = 0 3 0 4 0 5 0 6 0

The ordered pair (0, -1) has an open circle on it (on the graph) because the equation used to acquire that point states that u < 0. So all points on that line are valid until u = 0. So this set of equations makes a function; the range is {y | y < 0} and the zeroes are x > 3.

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can you explain why there is ordered pair (0,-1) when u < 0?

As the explanation stated, the grapher was taking points within the domain (or at the edge) in order to complete that portion of the graph. Because "u = 0" is not actually within that rule's domain, the point is not actually on that portion of the graph. However, how else would one know where to draw the line, if not by finding that edge-point? So the edge-point is found, according to the rule, but an open dot is drawn, indicating that the first rule's line goes up to, but does not quite include, the edge-point.

and why the range is {y|y ≤ 0} but not {y|-4 ≤ y ≤ 0}

If you've forgotten what linear functions (such as "y = x - 1") look like, then plot some more points for the first rule. That is, extend the graph. Then you should see that the graph does indeed go below the value y = -4.

 

[Harry_the_cat]

g(u)= {u-1 if u <0; u² -2u - 3 if 0 ≤ u ≤ 3; 0 if u > 3

can you explain why there is ordered pair (0,-1) when u < 0? and why the range is {y|y ≤ 0} but not {y|-4 ≤ y ≤ 0}

1st question: can you explain why there is ordered pair (0,-1) when u < 0?
You'll notice if you look closely at the graph that the point (0, -1) is an empty circle, meaning that this point is not included in the function, but every point leading up to it (along the line
g(u)=u-1 from the left) is included. You couldn't just look at -1 and stop there. What about -0.1, -0.01, -0.001 etc? So the line continues up to (but not including) the point (0, -1).
When x=0, the second part of the piecewise function applies. Note that (0, -3) is a coloured-in circle on the graph. So (0, -3) IS part of the graph whereas (0, -1) is NOT.

2nd question: Note that for the first part of the function, the domain is u<0. The table is just a guide to get the shape and see what's happening. Why should the table start at u=-2?
What about u=-3, u=-4, u=-100000? The line g(u)=u-1 continues downwards to the left indefinitely, hence the range as stated.

Hope that helps you understand it better!

see above

 

[Dale10101]

fab u

g(u)= {u-1 if u <0; u² -2u - 3 if 0 ≤ u ≤ 3; 0 if u > 3

can you explain why there is ordered pair (0,-1) when u < 0? and why the range is {y|y ≤ 0} but not {y|-4 ≤ y ≤ 0}

There is no ordered pair (0, -1) defined by y = u -1, u < 0.

On the graph you circle (0 -1) to show that it is NOT an ordered pair of that function but that every point on the continuous function that approaches (0, -1) from the left is.

The range of the the portion of the function y = u -1, u < 0 depicted on the graph is { y| -3 ≤ y ≤ 0 } however the range of the function itself is { y|y ≤ 0 }.