#color(blue)("Some observations")# The #x^2# is positive so the general graph shape is #uu# Consider the generalised form of #y=ax^2bxc# The #bx# part of the equation shifts the graph left or right You do not have any #bx# type of value in your equation So the graph is central about the yaxis The #c# part of the equation is of value 1 so it lifts the vertex upThis is a very quick review of how to graph y = sqrt(x) 3 with the 'radical dance' and by table For more information, go to bitly/graph_transfCos(x^2) (x−3)(x3) Zooming and Recentering To zoom, use the zoom slider To the left zooms in, to the right zooms out When you let go of the slider it goes back to the middle so you can zoom more
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X^2+(y-sqrt(x^2))^2=1 graph-Take image convolution image of X^2 (y 2)^2 1; To sketch the graph of y = sqrt(x 2 1), first sketch a graph of y = x 2, then translate upward by 1 unit to get y = x 2 1 Finally, sketch the graph of y = sqrt(x 2 1) by plotting the square root of the y values of the previous graph (y = x 2 1) Ie, instead of plotting (1, 2), plot (1, sqrt(2)) and so on Your final graph will be roughly similar to the graph of y = x 2 1 except
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#Attempt to plot equation x^2 y^2 == 1 import numpy as np import matplotlibpyplot as plt import math x = nplinspace(1, 1, 21) #generate nparray of X values 1 to 1 in 01 increments x_sq = i**2 for i in x y = mathsqrt(1(mathpow(i, 2))) for i in x #calculate y for each value in x y_sq = i**2 for i in y #Print for debuggingTangent of y=sqrt (x^21), (0, 1) \square! We can see that graph y = sqrt(x) using its inverse function y = x 2 with x greater than or equal to 0 Pretty handy, huh?
Algebra Graph y = square root of 1x^2 y = √1 − x2 y = 1 x 2 Find the domain for y = √1 −x2 y = 1 x 2 so that a list of x x values can be picked to find a list of points, which will help graphing the radical Tap for more stepsPlot of `y=sqrt(x)` is red in color and plot of `y=1/2x` is blue in color The curves intersect at `(0,0)` and `(4,2)` First let's find the area of the region,Graph x^2y^2=1 x2 y2 = 1 x 2 y 2 = 1 This is the form of a circle Use this form to determine the center and radius of the circle (x−h)2 (y−k)2 = r2 ( x h) 2 ( y k) 2 = r 2 Match the values in this circle to those of the standard form The variable r r represents the radius of the circle, h h represents the xoffset from the
Contact Pro Premium Expert Support »Swap sides so that all variable terms are on the left hand side \sqrt {x1}=y x 1 = y Square both sides of the equation Square both sides of the equation x1=y^ {2} x 1 = y 2 Subtract 1 from both sides of the equation Subtract 1 from both sides of the equationGraph `y=sqrt(x^21)` If `f(x)=sqrt(x^21)`, note that `f(x)=f(x)`, so the graph is symmetric about the yaxisIn other words, the graph has a mirror image over the yaxis So if you can graph
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$\begingroup$ Square both sides Your surface is a quadric surfaceThey are classified into 57 types (I don't remember exactly how many) As one of the answers states, yours is a hyperboloid of one sheetYour favorite calculus textbook should discuss this in detail and have picturesSolve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and moreFree PreAlgebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators stepbystep
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3 Answers3 Write it as x 2 z 2 = y 2 Note that y is the hypotenuse of a triangle with length x and height z So, this forms a circular cone opening as you increase in y or decrease in y This figure is the (double) cone of equation x 2 = y 2 − z 2 The gray plane is the plane ( x, y) You can see that it is a cone noting that for any yNote To see a different view, use the mouse to click and drag in the plot To change some of the plot options, rightclick on the plot, and select an option For example, try using the mouse to add axes to the plot Just like the plot command, the plot3d command can be given many optional arguments Look forCompute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history
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Explanation divide terms on numerator by x2 ⇒ y = x1 2 x2 x x2 = x− 3 2 x−1 now differentiate each term using the power rule dy dx = − 3 2 x− 5 2 − 1x−2 and writing with positive indices gives dy dx = −( 3 2x5 2 1 x2)To plot a function just type it into the function box Use "x" as the variable like this Examples sin(x) 2x−3;Divide 0 0 by 4 4 Multiply − 1 1 by 0 0 Add − 2 2 and 0 0 Substitute the values of a a, d d, and e e into the vertex form a ( x d) 2 e a ( x d) 2 e Set y y equal to the new right side Use the vertex form, y = a ( x − h) 2 k y = a ( x h) 2 k, to determine the values of a a, h h, and k k
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Get stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us! 1 I am trying to plot the following equation in MATLAB ratio = sqrt (11/ (kr)^2) With k and r on the x and y axes, and ratio on the z axis I used meshgrid to create a matrix with values for x and y varying from 1 to 10 x,y = meshgrid ( 1110, 1110);Solve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and more
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Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, historyThere are various techniques that can be employed to plot the graph of the function The most popular one is the use of the table of values (x) = \sqrt{ 1 x^2} {/eq}, we will use the tableWolfram Community forum discussion about How to plot 1/sqrt(x^2y^2z^2) in mathematica?
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Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers Visit Stack ExchangeMathz=\sqrt{4x^2y^2}\iff x^2y^2z^2=4\,,\,z\geq0/math This is the top half of a sphere of radius 2, centered at the originSolve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and more
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Active Oldest Votes 3 just plot the graph in the first quadrant and then flip it to all 4 quadrants you are guaranteed to get the accurate graph here's why the flipping works consider x y = 1 you can only plug in positive x's and output will be positive y's only now consider xThe equation is now solved x^ {2}x1=y Swap sides so that all variable terms are on the left hand side x^ {2}x=y1 Subtract 1 from both sides x^ {2}x\left (\frac {1} {2}\right)^ {2}=y1\left (\frac {1} {2}\right)^ {2} Divide 1, the coefficient of the x term, by 2 to get \frac {1} {2}Plot frac(X^2 (y 2)^2 1) directional derivative of X^2 (y 2)^2 1 in direction (1, 1) at point (2, 3) series X^2 (y 2)^2 1;
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Stay on top of important topics and build connections by joining Wolfram Community groups relevant to your interests how do i plot the section of a cone z = 9sqrt(x^2 y^2) in the cylinder of r=2 Follow 1 view (last 30 days) Show older comments Carlos Perez on Vote 1 ⋮ Vote 1 Commented John D'Errico on pretty much what the question says ive tried two different ways and none of them have worked I can post what i have if needed Plot y^2 = x^2 1 (The expression to the left of the equals sign is not a valid target for an assignment)
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Precalculus Graph x^2 (y1)^2=1 x2 (y − 1)2 = 1 x 2 ( y 1) 2 = 1 This is the form of a circle Use this form to determine the center and radius of the circle (x−h)2 (y−k)2 = r2 ( x h) 2 ( y k) 2 = r 2 Match the values in this circle to those of the standard form The variable r r represents the radius of the circle, h hFormula for love X^2(ysqrt(x^2))^2=1 (wolframalphacom) 2 points by carusen on hide past favorite 41 comments ck2 onGraph the parabola, y =x^21 by finding the turning point and using a table to find values for x and y
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Although Mark's answer is the "natural" one, here are other options just for completeness Use Plot3D, after performing a rotation Plot3D{1, 1} Sqrt1 x x, {x, 1, 1}, {y, 1, 1}, AspectRatio > 1Is y=sqrt {x^21} a function? How to plot x^2 y^2 = 1?
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The graph of mathx^2(y\sqrt3{x^2})^2=1/math is very interesting and is show below using desmosLearn more about plot MATLAB I note that your example has the wrong sign on the y^2 term, which is irrelevant, since your example is perfectly validQuora Something went wrong Wait a moment and try again
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Graph y=2^x y = 2x y = 2 x Exponential functions have a horizontal asymptote The equation of the horizontal asymptote is y = 0 y = 0Plot x^2 3y^2 z^2 = 1 WolframAlpha Rocket science?Free Hyperbola calculator Calculate Hyperbola center, axis, foci, vertices, eccentricity and asymptotes stepbystep
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Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history See below For the equation y=sqrt(x2) we can graph this starting with an understanding of the graph, sqrtx and adjusting from there Let's take a look at that graph first graph{sqrtx 1, 10, 3, 5} The graph of sqrtx starts at x=0, y=0 (since we're graphing in real numbers on the x and y axis, the value under the square root sign can't be negative) then passes through x=1, y=1 and xSquare both sides x 2 x y y = 1 Isolate the radical and square x y = ( 1 − x − y 2) 2 Rotate the axes by π / 4 to new axes X, Y Y 2 − X 2 2 = ( 1 − 2 Y 2) 2 Simplfying gives X 2 = 2 2 Y − 1, which is the equation for a parabola
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Have a question about using WolframAlpha? The drawing of the graph should then look something like graph{sqrt(1x^2) 1734, 187, 497, 1305} The graph is always concave up, has a minimum at (0,1), and switches from increasing to decreasing at x=0
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