determine whether the given points are on the graph of the equation. equation: y
ID: 3102880 • Letter: D
Question
determine whether the given points are on the graph of the equation. equation:y=x^3-2(square root x)
points: (0,0) (1,1) (1,-1)
list the intercepts and test for symetry
y^2 -x-4=0
y=x^2-4
______
2x^4
write the standard form of the equation and the generl form of the equationof each circle of radius r and center (h,k); graph each circle:
(h,kr=3; (h,k)=(1,0),r=4;(h,k)=(2,-3)
a.find the center (h,k)and the radius r of each circle; b. graph each circle; c. find te intercepts if any, of the graphs.
x^2 + y^2+4x+2y-20=0
find the general form of the equation center at the point 1,0 and containing the point -2,3
Explanation / Answer
determine whether the given points are on the graph of the equation. equation: y=x^3-2(square root x) points: (0,0) (1,1) (1,-1) Let x = 0. Then y = x^3 - 2sqrt(x) = 0^3 - 2(0) = 0 So, (0,0) is on the graph of the equation. Let x = 1. Then y = x^3 - 2sqrt(x) = 1^3 - 2(1) or 1^3 - 2(-1) = -1 or 3 So, (1,-1) is on the graph, but (1,1) is not. ___________________________________________________________ Vertical intercepts occur when x = 0, Horizontal intercepts occur when y = 0. y^2 -x-4=0 Let x = 0. y^2 -0-4=0 y^2 - 4=0 y^2 =4 y = 2 or -2 So, (0,2), (0,-2) are the vertical intercepts. Let y = 0. 0^2 -x-4=0 -x-4=0 x = -4 So, (0,-4) is the only horizontal intercept. Test for symmetry: solve for y. y = sqrt( x + 4) replace x with -x. y = sqrt(-x + 4) this is not -sqrt(x+4) and it is not sqrt(x+4) so it is not symmetric about the x axis or the origin. (Looking at a graph, you can see that it is symmetric about the x axis. Meaning y^2 - 4=x is the same as (-y)^2 - 4=x. But that isn't a common test for symmetry) ___________________________________________ (h,kr=3; and (h,k)=(2,-3) don't make sense, so I will do (h,k)=(1,0),r=4 the center is at (1,0) and the radius is 4. the general form is (x - h)^2 + (y - k)^2 = r^2 so, (x-1)^2 + y^2 = 16 or, (x-1)^2 + y^2 -16 = 0 To find the intercepts, find (x-h)^2 + (y-k)^2 - r^2 = 0 for both circles. Set the equations equal to each other. Then, solve the equations for x and y. Note - there may be multiple solutions. _______________________________________ We need to factor x^2 + y^2+4x+2y-20=0 Try (x + 2)^2 + (y + 1)^2. x ^2 + 4x + 4 + y^2 + 2y +1 x ^2 + 4x + y^2 + 2y + 5 x ^2 + 4x + y^2 + 2y + 5 - 25 15 = x^2 + y^2+4x+2y-20 (x + 2)^2 + (y + 1)^2 - 25 = x^2 + y^2+4x+2y-20=0 (h,k) = (-2,-1) r = 5 _________________________________________ (h,k) = (1,0) the distance between (1,0) and (-2,3) is the radius of the circle. So r = sqrt( (1--2)^2 + (0--3)^2) = sqrt( 9 + 9) = sqrt(18) Use the formula for standard form.
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