X^2 (y-x^(2/3))^2=1 graph 203897-X^2+(y-3 sqrt(x^2))^2=1 graph

Free functions calculator explore function domain, range, intercepts, extreme points and asymptotes stepbystep Explanation This is the equation of a circle with its centre at the origin Think of the axis as the sides of a triangle with the Hypotenuse being the line from the centre to the point on the circle By using Pythagoras you would end up with the equation given where the 4 is in fact r2 To obtain the plot points manipulate the equation as below Given x2 y2 = r2 → x2 y2 = 4 The description below represents function a and the table represents function b function a the function is 5 more than 3 times x function b x y −1 2 0 5 1 8 which statement is correct about the slope and yintercept of the two functions?

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X^2+(y-3 sqrt(x^2))^2=1 graph

X^2+(y-3 sqrt(x^2))^2=1 graph-Divide 2, the coefficient of the x term, by 2 to get 1 Then add the square of 1 to both sides of the equation This step makes the left hand side of the equation a perfect square x^ {2}2x1=\frac {y1} {3}1 Square 1 x^ {2}2x1=\frac {y2} {3} Add \frac {y1} {3} to 1 \left (x1\right)^ {2}=\frac {y2} {3}About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How works Test new features Press Copyright Contact us Creators

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I have no idea how this equation \\begin{equation} (x^2 y^2 1)^3 x^2 y^3 = 0 \\end{equation} Produces this picture Can someone provide a general explanation of plotting this function? Set y = 0 because the xaxis crosses the yaxis at y = 0 0 = 2(x 1)2 3 Subtract 3 from both sides −3 = 2(x 1)2 Divide both sides by 2 − 3 2 = (x 1)2 Square root both sides √− 3 2 = x 1 As we are square rooting a negative value it means that the curve does NOT cross or touch the xaxis If you want to graph on the TI specifically, you'll need to do some easy math x² (yx^ (2/3))² = 1 (y x^ (2/3))² = 1 x² y x^ (2/3) = ±√ (1 x²) y = x^ (2/3) ± √ (1 x²) Now you have Y in terms of X The TI series doesn't have a ± sign, though, so you'll need to graph the two equations as a list Type this in in Y

Free tangent line calculator find the equation of the tangent line given a point or the intercept stepbystepGraph the parabola, y =x^21 by finding the turning point and using a table to find values for x and yPopular Problems Algebra Graph y=2 (x1)^23 y = −2(x − 1)2 3 y = 2 ( x 1) 2 3 Find the properties of the given parabola Tap for more steps 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 a = − 2 a = 2 h = 1 h = 1

5 0 5 2 0 2 = 25 0 = 25 3 4 3 2 4 2 = 9 16 = 25 0 5 0Graph x/2 3Graph halfx 31) y = 2(x 2)^2 2 2) y = (x 2)^2 2 3) y = (x 2)^2 2 4) y = (x 2)^2 2 the answers to ihomeworkhelperscom

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The graph of y=x^22x1 is translated by the vector (2 3)The graph so obtained is reflected in the xaxis and finally it is stretched by a factor 2 parallel to yaxisFind the equation of the final graph is the form y=ax^2bxc, You can view more similar questions or ask a new questionExtended Keyboard Examples Upload Random Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, musicGraph (x3)^2 (y2)^2=4 (x − 3)2 (y − 2)2 = 4 ( x 3) 2 ( y 2) 2 = 4 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

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Question graph the equations y=3/2 x1 Answer by Fombitz () ( Show Source ) You can put this solution on YOUR website! 1) find the lowest point/ highest point of the equation 2) identify the convex of the equation 3) find any (as more as you can) point which are at the xaxis and the equation or at the yaxis and the eqution 4) draw~ Make equation easier to read y = 3(x 2)^2 1 = 3(x^2 (2)(x)(2) 2^2) 1 = 3(x^2 4x 4) 1 = 3x^2 12x 12 1 = 3x^2 12x 13 find the highest/lowestPlane x = 1 2 The trace in the x = 1 2 plane is the hyperbola y2 9 z2 4 = 1, shown below For problems 1415, sketch the indicated region 14 The region bounded below by z = p x 2 y and bounded above by z = 2 x2 y2 15 The region bounded below by 2z = x2 y2 and bounded above by z = y 7

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13 Surface 24x 24y2 9z = 35;Algebra questions and answers; Probably you can recognize it as the equation of a circle with radius r = 1 and center at the origin, (0,0) The general equation of the circle of radius r and center at (h,k) is (x −h)2 (y −k)2 = r2 Answer link

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Answer (1 of 6) That's a circle Compare x^2y^22x=0 with the general equation of a circle x^2y^22gx2fyc=0 and write down the values of the constants g, f and c g=1 f=0 c=0 The centre of the circle is (g,f)\equiv(1,0) The radius of the circle is \sqrt{g^2f^2c}=\sqrt{1} IfView interactive graph > Examples x^2y^2=1; Slope of tangent to curve at x = 2 is m \(=\frac{1\times47}{1^2}=\frac{47}{1}\) = 3 Ordinate of point P on curve when x = 2 is y \(=\frac{2^23}{21}\) = 7/1 = 7 Hence, point (2, 7) lies on curve Thus tangent of given curve has slope 3 and passing through point (2, 7) when x = 2 ∴ Equation of tangent at x = 2 is y y 1 = m(x x 1

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Incoming Term: x^2+(y-x^(2/3))^2=1 graph, x^2+(y-3 sqrt(x^2))^2=1 graph,