Is function symmetric w/respect to x-axis, y-axis or origin?
A function can not be symmetric with respect to the x-axis. This is because a function's graph would have to pass the vertical line test (one output for each input). To check for y-axis symmetry or origin symmetry, replace x with -x. Simplify as much as possible and see if you (a) get the same function back (y-axis symmetry) or (b) get exactly the negative of the original function (origin symmetry). A function need not have symmetry. It helps to know that if n is odd, then (-x)^n = -x^n and if n is even then (-x)^n = x^n. For your example: let x be replaced with -x y(-x) = (-x)^7 - (-x)^9 = -x^7 + x^9 = -(x^7 - x^9). So you get exactly the negative of the function. This function is odd (origin symmetry). If you have a relation---not necessarily a function---you can check for x-axis symmetry by replacing y with -y. You can use any legitimate algebra to simplify including distribution. *** Update ***** In response to your additional details: Your function IS symmetric with respect to the origin. When you put -x in for x, you get the same function with a minus sign. This is origin symmetry.
If a graph is symmetric with respect to the x axis?
that means it will look the same if you reflect it over the x axis. Think of points 0,4 and 0,-4; they are reflections of each other over the x axis, now we know how to do it! so the reflected point will be (-5,-8) !
What is a point that is symmetrical to (0,8) with respect to the y-axis?It’s on the y axis, so it’s its own such symmetry point.
Is y=2x^2-5 symmetric with respect to the x axis, y axis or origin. show work?
y = 2x^2 - 5 is a graph of parabola. ------------------------------------- When x = # , this is an equation of a VERTICAL line. Similarly, when "x" is squared, "y" is to the power of one, the parabola is a VERTICAL parabola. So, now, we have established that the above equation is a VERTICAL parabola. --------------------------------------... A VERTICAL parabola is symmetric with respect to the y-axis, regardless of which direction the parabola opens (up or down). --------------------------------------... ANSWER: This parabolic equation is symmetric with respect to the y-axis.
How do I test if an equation is symmetrical with respect to x or y axis or origin?
Now a touch definition first. a level ( x,y ) is asserted to be symmetric with recognize to the : a million) x -Axis if the point (x, -y) nonetheless lies on that curve. 2) y -Axis if the point (-x, y) nonetheless lies on that curve. 3) The starting place (0,0) if the point (-x, -y) nonetheless lies on that curve. So we only favor to 'verify' the situations (a million), (2) & (3) above for the equation given for our curve. to envision for (a million), evaluate 4(x)² - 9(-y)² = 4x² - 9y² = a million; and subsequently this aspect lies on the curve because it satisfies the equation! in addition, for ( 2 ), 4(-x)² - 9(y)² = 4x² - 9y² = a million; and subsequently this aspect lies on the curve because it satisfies the equation! For (3), 4(-x)² - 9(-y)² = 4x² - 9y² = a million; and subsequently this aspect lies on the curve because it satisfies the equation! subsequently we've shown that our graph for the curve 4x² - 9y² = a million is symmetrical for all 3 x - axis, y - axis and starting place! wish this helped. :)
The point may(3,4) lie on the curve, but the symmetry of the curve cannot explain about the point (3,4)Because, the curve symmetric to the x-axis means, the x-axis divides the curve on the Cartesian plane exactly into 2 halves.If the point on the curve falls on 1st quadrant (x,y) , then (x,-y)(4th quadrant) will also lie on the curve and vice versa. Similarly, If the point on the curve falls on 2nd quadrant (-x,y) , then (-x,-y)(3rd quadrant) will also lie on the curve and vice versa.Similarly,if the curve symmetric to the y-axis means, the y-axis divides the curve on the Cartesian plane exactly into 2 halves.If the point on the curve falls on 1st quadrant (x,y) , then (-x,y)(2nd quadrant) will also lie on the curve and vice versa. Similarly, If the point on the curve falls on 3rd quadrant (-x,-y) , then (x,-y)(4th quadrant) will also lie on the curve and vice versa.
You have got a function [math]f(\theta)[/math]You want to check whether this function is symmetric with respect to the polar axis, i.e. with respect to the line [math]\theta = 0,[/math] i.e. the Cartesian X axis.Just take the mirror image of the function with respect to the X axis and see whether the values are the same.To do this in this coordinate system, replace [math]\theta[/math] in the equation with [math]-\theta[/math] i.e. check if [math]f(\theta)=f(− \theta).[/math]If the value of the function is the same then the function is symmetric with respect to the polar axis.In this specific case, [math]f(\theta )= 4\sin 2\theta[/math][math]\Rightarrow \qquad f(-\theta )= 4\sin (-2\theta ) [/math][math]= -4\sin 2\theta = -f(\theta )[/math]Thus we see that [math]f(\theta ) \ne f(-\theta ),[/math] which implies that the function is not symmetric about the polar axis.In general, even functions (in the polar coordinate system) have the property of symmetry about the polar axis, i.e. the X axis .
Symmetry of a trig function?
If you fold it around the y axis, the point (90,1), which is on the graph, moves to (-90,1). (My brain is in degree mode, if you can't tell.) But unfortunately, sin -90 is actually -1, not +1. Going 90 degrees backwards (clockwise) on the unit circle takes you to the negative y-axis. The sine function is odd: f(-x) = -f(x). sin (-x) = -sin(x). Odd functions are symmetric about the origin. The cosine function is even: f(-x) = f(x). cos (-x) = cos(x). Even functions are symmetric about the y-axis. The easiest way to remember your even functions from your odd functions is to compare them to f(x) = x^2 and g(x) = x^3. f is even, symmetric about the y-axis, and has an EVEN exponent. g(x) is odd, symmetric about the origin, and has an ODD exponent. The names 'even' and 'odd' may make no sense now, but they will in calculus. If you use a polynomial with a lot of terms to ESTIMATE sin x at x=0, the "best" polynomial you get will only have odd powers of x. And if you do the same for cosine, you only get even powers of x.
Consider any point in the graph, say [math](a, b)[/math]Symmetry about x-axis means the point [math](a, -b)[/math]is in the graph too. Eg.: [math]x = y^2[/math]Here x-axis acts as the mirror. Note that the given example does not satisfy the requirement of a function for [math]y = f(x)[/math]Symmetry about y-axis means the point [math](-a, b)[/math]is in the graph too. Eg.: [math]y = x^2[/math]Here y-axis acts as the mirror.Symmetry about Origin means the point [math](-a, -b)[/math]is in the graph too. Here Origin is the point mirror or we have rotational symmetry [math](\theta < 360 degrees)[/math]Now about the difference:When there is symmetry with respect to both x and y axis[math](a, b)\rightarrow [/math](x-axis symmetry)[math]\rightarrow (a, -b)\rightarrow [/math](y-axis symmetry)[math]\rightarrow (-a, -b)\rightarrow [/math]origin symmetry[math](a, b)\rightarrow [/math](y-axis symmetry)[math]\rightarrow (-a, b)\rightarrow [/math](x-axis symmetry)[math]\rightarrow (-a, -b)\rightarrow [/math]origin symmetryi.e. when there is symmetry with respect to x and y axes, the point [math](a, b)\rightarrow (a, -b),(-a, b)[/math] and [math] (-a, -b)[/math], thus leading to origin symmetry.But Origin symmetry need not lead to symmetry about x and y axes!i.e. [math](a, b)\rightarrow [/math]origin symmetry[math] \rightarrow (-a, -b)[/math]Check this:[math]x^2+y^2 = 1[/math] is the equation of the unit circle about the origin, clearly it is symmetrical about the x and y axes and about the origin.But [math]y = x^3[/math] is symmetric about the origin and NOT symmetric about x and y axes.Symmetry about x and y axes implies Symmetry about the origin, butSymmetry about the origin does not imply Symmetry about x and y axes.Hope this helped! :)Thanks for the question!