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askor

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askor

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Have you done any research on this? What have you found?

- #3

Mayhem

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- #4

FactChecker

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- #5

askor

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I only know how to draw a graph of quadratic function. This function has a power of 3.

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Where are the turning points? What is the value of the function at those points? How does the function behave as ##x \to \pm \infty##?

I only know how to draw a graph of quadratic function. This function has a power of 3.

- #7

berkeman

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Google search forHow do you draw the graph of

https://www.google.com/search?q=algebra+how+to+draw+graph&ie=utf-8&oe=utf-8&client=firefox-b-1

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We call it curve discussion here at school. The recipe is always the same:What math the drawing a graph, for example: y = x^2, is taught?

- Determine the zeros of the function.
- Differentiate as often as it is possible.
- Determine all zeros and sign changes of all derivatives.
- Determine all poles.
- Determine the asymptotes at the poles.
- Determine the asymptotes at infinity.

- #9

FactChecker

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- #10

askor

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Yes, I learned calculus of differentiation and integration.

- #11

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Then you should be able to do the steps that @fresh_42 put in the post above (possibly with some review). His list looks complete to me.Yes, I learned calculus of differentiation and integration.

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- #13

mathwonk

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if the graph of a differentiable function changes direction, then the derivative equals zero somewhere in between. This means the graph does not change direction between any two consecutive zeroes of the derivative. Thus the most efficient way to graph a function is to plot the zeroes of the drivative and then to determine the limits at ± infinity. If the derivative has no zeroes, and even if it does, you will also want to plot a few other points, such as where it crosses the x-axis and/or y axis, and where it changes curvature. It is thus useful to plot zeroes of the second derivative, since curvature can only change at these points.

It follows that efficient graphing can only be taught in or after a course which teaches derivatives, hence usually calculus, at least today in the US. Note however that over 100 years ago in the US, we had more elementary books such as Treatise on Algebra by Charles Smith (1888), that did treat derivatives, so my father may have learned this in high school in the early 1900's.

It follows that efficient graphing can only be taught in or after a course which teaches derivatives, hence usually calculus, at least today in the US. Note however that over 100 years ago in the US, we had more elementary books such as Treatise on Algebra by Charles Smith (1888), that did treat derivatives, so my father may have learned this in high school in the early 1900's.

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symbolipoint

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THAT, or at least mostly.

Students learn to make simple graphs in "Algebra 1", and the studies for making graphs continue from there.

- #15

symbolipoint

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If at least you have studied your Algebra 1, then you know to either (1) use a graphing tool, or (2) plug in values for x and evaluate corresponding y values; and then plot your points and draw your curve. Too simple? You learn more about such graphs in College Algebra.

I only know how to draw a graph of quadratic function. This function has a power of 3.

- #16

askor

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How do I draw the V shear and M moment graph like above?

- #17

symbolipoint

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- #18

mathwonk

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I will go out on a limb and try to answer the original question, but I could easily make mistakes.

To graph y = 9x - x^3/9, first set y = 0 and get x = 0, 9, -9. If one also notes that the leading term is a constant multiple of -x^3, one knows the graph comes down from + infinity on the extreme left and goes down to - infinity at the extreme right. Hence just graphing these 3 points, (-9,0), (0,0), and (9,0), one knows the graph comes down from the left, crosses the x-axis at (-9,0), then at some point between x= - 9, and x= 0 , the graph turns up again, crossing the x-axis at (0,0), continues upward, then turns down again at some point between x=0 and x = 9, and goes down again, crossing the x-axis at (9,0), then continues on down to minus infinity. Calculus is not really needed for this rough graph, which looks a bit like a fishhook on the left, and an upside down fishhook on the right, with their points meeting at (0,0).

To find the 2 turning points, use calculus to take a derivative, getting y' = 9 - x^2/3, and set it equal to zero, getting x^2 = 27, or x = ± sqrt(27). Plotting these two points gives the turning points as (-sqrt(27), -6sqrt(27)), and (sqrt(27), 6 sqrt(27)).

To clarify that the curvature does indeed change from concave up to concave down at (0,0), compute the second derivative y'' = -2x/3, set it equal to zero, getting x=0.

To compute the slope of the graph at this "inflection" point, compute y'(0) = 9, getting the slope of the tangent line at (0,0).

That is about what a calculus student of mine would learn.

To graph y = 9x - x^3/9, first set y = 0 and get x = 0, 9, -9. If one also notes that the leading term is a constant multiple of -x^3, one knows the graph comes down from + infinity on the extreme left and goes down to - infinity at the extreme right. Hence just graphing these 3 points, (-9,0), (0,0), and (9,0), one knows the graph comes down from the left, crosses the x-axis at (-9,0), then at some point between x= - 9, and x= 0 , the graph turns up again, crossing the x-axis at (0,0), continues upward, then turns down again at some point between x=0 and x = 9, and goes down again, crossing the x-axis at (9,0), then continues on down to minus infinity. Calculus is not really needed for this rough graph, which looks a bit like a fishhook on the left, and an upside down fishhook on the right, with their points meeting at (0,0).

To find the 2 turning points, use calculus to take a derivative, getting y' = 9 - x^2/3, and set it equal to zero, getting x^2 = 27, or x = ± sqrt(27). Plotting these two points gives the turning points as (-sqrt(27), -6sqrt(27)), and (sqrt(27), 6 sqrt(27)).

To clarify that the curvature does indeed change from concave up to concave down at (0,0), compute the second derivative y'' = -2x/3, set it equal to zero, getting x=0.

To compute the slope of the graph at this "inflection" point, compute y'(0) = 9, getting the slope of the tangent line at (0,0).

That is about what a calculus student of mine would learn.

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- #19

symbolipoint

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One usually learns to graph polynomial functions, which your example here resembles, in College Algebra.

I only know how to draw a graph of quadratic function. This function has a power of 3.

I said, "resembles" not "is". Some people will tell that a polynomial function has integer coefficients. I am not fully certain if this is or is not correct.

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Polynomials can have coefficients from any ring or field.I said, "resembles" not "is". Some people will tell that a polynomial function has integer coefficients. I am not fully certain if this is or is not correct.

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MidgetDwarf

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- #22

MidgetDwarf

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Only the exponents of the variables of a polynomial need to be positive integers (including 0).One usually learns to graph polynomial functions, which your example here resembles, in College Algebra.

I said, "resembles" not "is". Some people will tell that a polynomial function has integer coefficients. I am not fully certain if this is or is not correct.

- #23

symbolipoint

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I made a technical word choice mistake. The exponents in polynomials are the Whole numbers. "Integers" obviously the wrong word choice.Only the exponents of the variables of a polynomial need to be positive integers (including 0).

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symbolipoint

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