Tutors Answer Your Questions about Graphs (FREE)
Question 916126: Draw the graphs of each of the following on the coordinate axes below:
f(x) = 12x^3 - 12x^2 - 24x
f(x) = (1/64)(x-3)^2(x+4)^3
Please explain in detail how I would figure these out/set them up. I am so lost right now I don't know where to start....
Thank you
Click here to see answer by MathLover1(20849)   |
Question 916232: How to set up a graph for this polynomial. I don't know where to start. Is there a way that I can know what the general shape of the graph will be before I start? I really need help
f(x) = (1/64)(x-3)^2(x+4)^3
Thank you
Click here to see answer by stanbon(75887) |
Question 916232: How to set up a graph for this polynomial. I don't know where to start. Is there a way that I can know what the general shape of the graph will be before I start? I really need help
f(x) = (1/64)(x-3)^2(x+4)^3
Thank you
Click here to see answer by MathLover1(20849) |
Question 916232: How to set up a graph for this polynomial. I don't know where to start. Is there a way that I can know what the general shape of the graph will be before I start? I really need help
f(x) = (1/64)(x-3)^2(x+4)^3
Thank you
Click here to see answer by Alan3354(69443)  |
Question 916237: How to graph polynomials. Where do I start? I am lost on these two problems. I don't know what I need to do in order to figure out the graph of:
f(x) = 4 log(base 4)(x + 8)
and
h(x) = (2x+6)(x–3)^2 (x+1)^3
Is there any way to tell what a graph will look like before starting?
Thank you in advance!
Click here to see answer by stanbon(75887) |
Question 916249: Understanding this graph:
http://i.imgur.com/nlOsmHN.png
What I don't understand is why does the graph approach from those sides? why not the opposite? why not approach from the more-negative side of -2 and the positive side of 8? I just I don't remember the rules for this.
Please help
Thank you
Click here to see answer by rothauserc(4718)  |
Question 916411: understand almost 90% of how to graph this but there are a few things I don't quiet understand.
h(x) = (2x+6)(x-3)^2(x+1)^3
y-int = (0,54)
x-int = (-3,0)(3,0)(-1,0)
My question is how do I determine what the degree is? and what the leading coefficient is?
Also, I am having confusion on what exactly a tangent is can it be thought of like a "straight line" or "the flat spot of a graph on the x-axis"
When I go to graph I understand all the parts except where there there is a flat spot for the x-intercept (-1,0) since it's an odd degree factor shouldn't it just go straight through it and up to the y-intercept? I don't understand the idea of the flattened spots when graphing.
It would be greatly appreciated if someone could explain this.
Thank you!
Click here to see answer by stanbon(75887) |
Question 916411: understand almost 90% of how to graph this but there are a few things I don't quiet understand.
h(x) = (2x+6)(x-3)^2(x+1)^3
y-int = (0,54)
x-int = (-3,0)(3,0)(-1,0)
My question is how do I determine what the degree is? and what the leading coefficient is?
Also, I am having confusion on what exactly a tangent is can it be thought of like a "straight line" or "the flat spot of a graph on the x-axis"
When I go to graph I understand all the parts except where there there is a flat spot for the x-intercept (-1,0) since it's an odd degree factor shouldn't it just go straight through it and up to the y-intercept? I don't understand the idea of the flattened spots when graphing.
It would be greatly appreciated if someone could explain this.
Thank you!
Click here to see answer by MathLover1(20849) |
Question 916560: Write the polynomial function P (lowest-possible degree) that could have the graph shown on the right
Finding the leading coefficient
http://i.imgur.com/ERfGDlj.png
I understand that it must be 4 degrees because it goes through at 3 bounces at -1 and through at at 2.
and it would be set up: (x+3)(x+1)^2(x-2)
I know it must be negative because it is downwards but how do I find the leading coefficient?
Thank you
Click here to see answer by MathLover1(20849) |
Question 916554: Figuring out the graph for:
f(x) = 12x^3 - 12x^2 - 24x
I understand that I must factor:
f(x) = 12x(x^2 - x - 2)
f(x) = 12x(x+1)(x-2)
x-int:(0,0)(-1,0)(2,0)
y-int:(0,0)
then I don't know what to do I have the points on the graph but don't know how I would graph this. Please help!
Click here to see answer by ewatrrr(24785)  |
Question 916554: Figuring out the graph for:
f(x) = 12x^3 - 12x^2 - 24x
I understand that I must factor:
f(x) = 12x(x^2 - x - 2)
f(x) = 12x(x+1)(x-2)
x-int:(0,0)(-1,0)(2,0)
y-int:(0,0)
then I don't know what to do I have the points on the graph but don't know how I would graph this. Please help!
Click here to see answer by Alan3354(69443) |
Question 916845: Figuring out graphing of this polynomial:
f(x) = 12x^3 - 12x^2 - 24x
what I have so far:
I know I can factor it down, therefore 12x(x+1)(x-2)
and the x-int would be (-1,0),(2,0)
But why is (0,0) also a factor? I don't understand that.
also how would I find the degree and leading coefficient? very lost on that concept
And when I plug this poly into a graphing calculator it has a steepness and a lowness at two points. I am in precalculus so I don't know the finer details of graphing/derivatives and I must be able to do this freehand how would I know how far up to make the graph and how far low to make the graph? Am I just suppose to estimate them for precalculus? What if both sides bounces how do I tell which one is higher or lower or if they are both the same?
Thank you
Click here to see answer by josgarithmetic(39617) |
Question 916833: Figuring out graphing of this polynomial:
f(x) = 12x^3 - 12x^2 - 24x
what I have so far:
I know I can factor it down, therefore 12x(x+1)(x-2)
and the x-int would be (-1,0),(2,0)
But why is (0,0) also a factor? I don't understand that.
also how would I find the degree and leading coefficient? very lost on that concept
And when I plug this poly into a graphing calculator it has a steepness and a lowness at two points. I am in precalculus so I don't know the finer details of graphing/derivatives and I must be able to do this freehand how would I know how far up to make the graph and how far low to make the graph?
Thank you
Click here to see answer by josgarithmetic(39617) |
Question 916872: How to find the leading coefficient with graph provided?
http://i.imgur.com/hslfSWo.png
I know that it has 4 degrees and it looks like: P(x)=(x+3)(x+1)^2(x-2) but I just cant seem to figure out what the leading coefficient is I know it's negative since the graphing is downwards, does it have something to do with the y-intercept?
Please help!
Thanks
Click here to see answer by solver91311(24713)  |
Question 916872: How to find the leading coefficient with graph provided?
http://i.imgur.com/hslfSWo.png
I know that it has 4 degrees and it looks like: P(x)=(x+3)(x+1)^2(x-2) but I just cant seem to figure out what the leading coefficient is I know it's negative since the graphing is downwards, does it have something to do with the y-intercept?
Please help!
Thanks
Click here to see answer by MathLover1(20849) |
Question 916856: Let h(x) = log(base 3) (x+9)
Find the domain of h.
On your graph show any asymptote with a dotted line labeled with its equation or intercept. label the intercepts with their coordinates or values.
Please explain how this solved
Thank you
Click here to see answer by solver91311(24713) |
Question 916902: Understanding why the graph looks the way it does:
f(x) = (4(x-5))/((x+2)(10-x))
I understand how to find the vertical/horizontal asymptotes and x and y intercept and I understand the middle line going through the middle of the graph but the other two lines curving along the asymptotes why are they there and how do I predict them and know when/where to plot them when I graph on paper? why are they in those positions and why not others?
http://i.imgur.com/Fkl11xT.png
Thank you
Please explain
Click here to see answer by stanbon(75887) |
Question 917137: How to find the leading coefficient and degree of:
f(x) = 12x^3 - 12x^2 - 24x
what I have:
f(x) = 12x(x^2 - x - 2)
f(x) = 12x(x+1)(x-2)
x-int: (-1,0),(2,0),(0,0)
y-int: (0,0)
Wouldn't the degree be 6 because you add the exponents so 3+2+1 = 6
The leading coefficient is 12 because 12 has x^3 attached to it which is the highest degree term.
When I plug this graph into a graphing calculator I get a graph pointing in different directions, so I know it must be negative.
What am I misunderstanding?
Thank you
Click here to see answer by stanbon(75887) |
Question 917137: How to find the leading coefficient and degree of:
f(x) = 12x^3 - 12x^2 - 24x
what I have:
f(x) = 12x(x^2 - x - 2)
f(x) = 12x(x+1)(x-2)
x-int: (-1,0),(2,0),(0,0)
y-int: (0,0)
Wouldn't the degree be 6 because you add the exponents so 3+2+1 = 6
The leading coefficient is 12 because 12 has x^3 attached to it which is the highest degree term.
When I plug this graph into a graphing calculator I get a graph pointing in different directions, so I know it must be negative.
What am I misunderstanding?
Thank you
Click here to see answer by MathLover1(20849) |
Question 917130: Rational functions end behavior and multiplicity?
I have the graph for:
(4(x-5))/(x+2)(10-x)
I understand how to plot the middle piece. But I am having trouble understanding the end behaviors on the regions outside of the vertical asymptotes. I read that when a rational function has an odd multiplicity/exponent it therefore approaching from the first and third quadrants. But this rational function I have is doing the exact opposite of that despite it have multiplicity of 1's in the bottom. What am I misunderstanding here?
http://i.imgur.com/mjjOGcD.png
please explain
Thanks
Click here to see answer by stanbon(75887) |
Question 917165: Really confused here with this logarithmic graph
h(x) = 4 log(base 4) (x+8)
What do I have to do in order to figure out the x and y intercepts and any other information in order to graph this? could someone show what needs to be done to graph this?
Is there a vertical/horizontal asymptote in this problem? what does the the 4 on the outside of the log do? I have no clue on how to figure this out
Please help
Click here to see answer by ewatrrr(24785) |
Question 917230: R(t) = 8 - (3)/(2x+4)
Please explain how I would go about finding the graph for this. Please give as many details as possible.
I know that the vertical asymptote is x=-2, and I guess there is no horizontal asymptote because there is no x's in the numerator.
How would I start finding points on the x-axis and overall figuring this out?
Thanks!
Click here to see answer by jim_thompson5910(35256) |
Question 917246: Please explain how I would determine the middle point on this graph and end behavior of the rational function:
f(x) = (8x-4x^2)/((x+2)^2)
What I know so far from this rational function is:
Vertical asymptote: x = -2
Horizontal asymptote: y = -4
X-intercept: (2,0),(0,0)
Y-intercept: (0,0)
Here is the graph in it's entirety: http://i.imgur.com/RIr1udF.png
What I want to know is how can I w/o a calculator figure out the middle point on the graph, I have (0,0) and (2,0) and understand that fine but how do I find that middle one freehand?
How do I know when the curve at the 3rd quadrant will appear? I don't understand why that is there.
Please explain as much as possible.
Thanks!
Click here to see answer by josgarithmetic(39617) |
Question 917221: How do I figure out the graph for: f(x) = 3(2^x)-6
I know that x=-6 is the horizontal asymptote
when I go to make a t-table my points are all over the place and not consistent.
like plugging in x=1 results in 0. x=2 results in 6...
What is the proper way to figure out this graph?
Thank you
Click here to see answer by josgarithmetic(39617) |
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