Steve S. answered • 01/23/14
Tutor
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Tutoring in Precalculus, Trig, and Differential Calculus
1.
y = -10/x
If x = 0 the y is undefined.
If x = 1/10^6, a very small number, y = -10/(1/10^6) = -10*10^6/1 = -10^7.
That's a very large negative number, so the graph plunges to negative values just to the right of the y-axis.
If x = -1/10^6, y = -10/(-1/10^6) = 10*10^6/1 = 10^7.
That's a very large positive number, so the graph shoots up to positive values just to the left of the y-axis.
That's a very large positive number, so the graph shoots up to positive values just to the left of the y-axis.
If x = 1, y = -10/1 = -10, so point (1,-10) is on the graph.
If x = -1, y = -10/(-1) = 10, so point (-1,10) is on the graph.
If x = 10^6, y = -10/10^6 = -10^(-5). So way out to the right the graph is just a little bit below the x-axis.
If x = -10^6, y = -10/(-10^6) = 10^(-5). So way out to the left the graph is just a little bit above the x-axis.
The axes are asymptotes of the graph because the graph continues to approach them but never touches them.
Both the domain and the range of the function is All Real Numbers except 0.
2.
y = 1/(x + 3) + 3
This is a translation of the function y = 1/x (which you should be able to analyze and graph now with your knowledge of problem 1) by shifting left 3 and up 3.
Good luck.
===== 4/10/14
2. How do I graph f(x) = 1/(x+3) + 3 and find the domain and range?
If you translate the graph of f(x) = 1/(x+3) + 3 right 3 and down 3 you get the graph of g(x) = 1/x.
Let’s analyze y = g(x) = 1/x; afterwards we can translate it back to get f(x).
Can’t divide by zero so x ≠ 0 is domain;
or all real numbers except 0;
or x e (-∞,0)U(0,∞).
Solving for x, x = 1/y.
Can’t divide by zero so y ≠ 0 is range;
or all real numbers except 0;
or y e (-∞,0)U(0,∞).
x = 0, the y-axis, is a vertical asymptote and
y = 0, the x-axis, is a horizontal asymptote.
An asymptote is a line that the function’s graph keeps getting closer to but never touches as x changes.
To see which side of an asymptote the function is on, substitute very small numbers or very large numbers:
0+ = 0 + 1/10^6 = 10^(-6)
0– = 0 - 1/10^6 = –10^(-6)
y(0+) = 1/(10^(-6)) = 10^6; a little to the right of the vertical asymptote y(x) is a very large positive number.
y(0–) = 1/(–10^(-6)) = –10^6; a little to the left of the vertical asymptote y(x) is a very large negative number.
y(10^6) = 1/(10^6) = +10^(-6); way out to the right y(x) is a little above the x-axis.
y(–10^6) = 1/(–10^6) = –10^(-6); way out to the left y(x) is a little below the x-axis.
The points (1,1) and (-1,-1) are on y = g(x) = 1/x. Graph the points and draw smooth curves through them in Quadrants I and III, being careful to approach but not touch the asymptotes. Your graph should look like this: http://tccl.rit.albany.edu/knilt/images/6/64/Hyperbola.gif.
Now, if we translate g(x) left 3 and up 3 we
If you translate the graph of f(x) = 1/(x+3) + 3 right 3 and down 3 you get the graph of g(x) = 1/x.
Let’s analyze y = g(x) = 1/x; afterwards we can translate it back to get f(x).
Can’t divide by zero so x ≠ 0 is domain;
or all real numbers except 0;
or x e (-∞,0)U(0,∞).
Solving for x, x = 1/y.
Can’t divide by zero so y ≠ 0 is range;
or all real numbers except 0;
or y e (-∞,0)U(0,∞).
x = 0, the y-axis, is a vertical asymptote and
y = 0, the x-axis, is a horizontal asymptote.
An asymptote is a line that the function’s graph keeps getting closer to but never touches as x changes.
To see which side of an asymptote the function is on, substitute very small numbers or very large numbers:
0+ = 0 + 1/10^6 = 10^(-6)
0– = 0 - 1/10^6 = –10^(-6)
y(0+) = 1/(10^(-6)) = 10^6; a little to the right of the vertical asymptote y(x) is a very large positive number.
y(0–) = 1/(–10^(-6)) = –10^6; a little to the left of the vertical asymptote y(x) is a very large negative number.
y(10^6) = 1/(10^6) = +10^(-6); way out to the right y(x) is a little above the x-axis.
y(–10^6) = 1/(–10^6) = –10^(-6); way out to the left y(x) is a little below the x-axis.
The points (1,1) and (-1,-1) are on y = g(x) = 1/x. Graph the points and draw smooth curves through them in Quadrants I and III, being careful to approach but not touch the asymptotes. Your graph should look like this: http://tccl.rit.albany.edu/knilt/images/6/64/Hyperbola.gif.
Now, if we translate g(x) left 3 and up 3 we
get f(x) = 1/(x+3) + 3.
How do we do that on graph paper?
First draw dashed lines x = –3 and y = 3. Treat these as translated y and x axes respectively. The translated origin is (–3,3).
Now draw g(x) on the translated axes; that graph is also f(x) relative to the regular axes.
The asymptotes of f(x) will be the translated asymptotes of g(x); i.e.,
How do we do that on graph paper?
First draw dashed lines x = –3 and y = 3. Treat these as translated y and x axes respectively. The translated origin is (–3,3).
Now draw g(x) on the translated axes; that graph is also f(x) relative to the regular axes.
The asymptotes of f(x) will be the translated asymptotes of g(x); i.e.,
vertical asymptote is x = –3 and
horizontal asymptote is y = 3.
The Domain of f(x) is x ≠ –3
and the Range is y ≠ 3.
The Domain of f(x) is x ≠ –3
and the Range is y ≠ 3.
Aubree S.
Hi. I have this same problem and I understand why your answer is right and everything, but if you take out the concept of asymptotes and limits could you explain how you know the range is just 0 just by looking at or manipulating the equation in a purely Albegra 1 kind of way?
Thanks.
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04/10/14