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-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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Steps for Solving Linear Equation y = f ( x ) = 2 ( x 1 ) y = f ( x) = 2 ( x 1) Use the distributive property to multiply 2 by x1 Use the distributive property to multiply 2 by x 1 y=2x2 y = 2 x 2 Swap sides so that all variable terms are on the left hand side From the given equation #x^2y^22x3=0# perform completing the square method to determine if its a circle, ellipse, hyperbola There are 2 second degree terms so we are sure it is not parabola #x^2y^22x3=0# #x^22xy^2=3# add 1 to both sides of the equation #x^22x1y^2=31# #(x^22x1)y^2=4# #(x1)^2(y0)^2=2^2# it takes the form #(xh)^2(yk)^2=r^2#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
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PreAlgebra Graph x^2y^2=1 x2 − y2 = −1 x 2 y 2 = 1 Find the standard form of the hyperbola Tap for more steps Flip the sign on each term of the equation so the term on the right side is positive − x 2 y 2 = 1 x 2 y 2 = 1 Simplify each term in the equation in order to set the right side equal to 1 1Answer (1 of 3) It's the equation of sphere The general equation of sphere looks like (xx_0)^2(yy_0)^2(zz_0)^2=a^2 Where (x_0,y_0,z_0) is the centre of the circle and a is the radious of the circle It's graph looks like Credits This 3D Graph isGraph a function by translating the parent function
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We identify from the equation A close to one big was to wonder or three See Question three request rule minus it equals Do it in the room three Africa list Do ditto Now goat off placeFor they gave any patient We have to draw the graph given equation is excess when less you wonder or three x way? Example 3 y = x 2 3 The "plus 3" means we need to add 3 to all the yvalues that we got for the basic curve y = x 2 The resulting curve is 3 units higher than y = x 2 Note that the vertex of the curve is at (0, 3) on the yaxis Next we see how to move a curve left and right Example 4 y = (x − 1) 2
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Circle on a Graph Let us put a circle of radius 5 on a graph Now let's work out exactly where all the points are We make a rightangled triangle And then use Pythagoras x 2 y 2 = 5 2 There are an infinite number of those points, here are some examples x y x 2 y 2;Yintercept of (0,1) Slope of (3/2) Slope= (Change in y)/ (Change in x)=3/2 Starting at the yintercept (0,1) Next x point 02=2 Next y point13=2Easy as pi (e) Unlock StepbyStep plot x^2y^2x Natural Language Math Input NEW Use textbook math notation to enter your math Try it × Extended Keyboard
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Steps to graph x^2 y^2 = 4Unlock StepbyStep (x^2y^21)^3=x^2y^3 Natural Language Math Input NEW Use textbook math notation to enter your mathCircleequationcalculator center (x2)^2(y3)^2=16 en Related Symbolab blog posts Practice, practice, practice Math can be an intimidating subject Each new topic we learn has symbols and problems we
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x^2 y^2 == 1 x^2y^2=1 It is a hyperbola, WolframAlpha is verry helpfull for first findings, The Documentation Center (hit F1) is helpfull as well, see Function Visualization, Plot3Dx^2 y^2 == 1, {x, 5, 5}, {y, 5, 5} ContourPlot3Dx^2 y^2 == 1, {x, 5, 5}, {y, 5, 5}, {z,Subtract 3 from both sides x^ {2}2x=y3 Multiply 2 and 1 to get 2 x^ {2}2x1=y31 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=y2 Add 3y to 1Square minus eight X less it and the road three Why is it was to Jiro?
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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 You can clickanddrag to move the graph around If you just clickandrelease (without moving), then the spot you clicked on will be the new center To reset the zoom to the original clickAnswer (1 of 9) Given , y = (x2)(x3)(x1) = x^{3} 4x^{2} x 6 Now , At x = 0 , y = 6 At y = 0 , x = 1 , 2 , 3 So the graph of the given equation goes through the points (0,6)(1,0),(2,0),(3,0) So the points in the graph will be like this — Now how do the points will connect to eacSubtracting x 2 from itself leaves 0 \left (y\sqrt 3 {x}\right)^ {2}=1x^ {2} ( y 3 x ) 2 = 1 − x 2 Take the square root of both sides of the equation Take the square root of both sides of the equation y\sqrt 3 {x}=\sqrt {1x^ {2}} y\sqrt 3 {x}=\sqrt {1x^ {2}} y 3 x = 1 − x 2 y 3 x = − 1 − x 2
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Answered 4 years ago Author has 76 answers and 12M answer views I am assuming you meant the following function x 2 ( y − x ( 3 / 2)) 2 = 1 also since you just wanted the graph and no explanation , So here it is you can easily graphX^2(y(x^2)^(1/3))^2 = 1 Natural Language;Question 7 (2 points) (Sec 25) The graph of y = f(x) is given below 4 y=f(x) 3 2 4 3 1 2 꾸 2 3 4 Use the
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WayneDeguMan Vertical asymptotes occur when the doniminator is zero ie when \displaystyle{2}{x}^{{2}}{3}{x}{2}={0} or, (2x 1)(x 2) = 0 HenceAnswer (1 of 3) The same way you plot anything Even with this equation being complicated looking, just assume that this elliptical mapping has some yvalue(s) for whatever xvalue(s) Since this is second order, we can expect it to have some values So, start off by making a list When x=0, yExtended 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, music
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Trigonometry Graph y= (x1)^32 y = (x − 1)3 − 2 y = ( x 1) 3 2 Find the point at x = −1 x = 1 Tap for more steps Replace the variable x x with − 1 1 in the expression f ( − 1) = ( ( − 1) − 1) 3 − 2 f ( 1) = ( ( 1) 1) 3 2 Simplify the result Tap for more stepsPiece of cake Unlock StepbyStep plot 3x^22xyy^2=1 Natural Language Math Input NEW Use textbook math notation to enter your math Try it × Extended Keyboard🔴 Answer 2 🔴 on a question Which equation is correct for the graph shown?
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The graph of y=x^2 is reflected in the xaxis, then stretched vertically by a factor of 2, and then translated 3 units to the left and 1 unit down I need to write the equation of this parabola in standard and general form
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