SOLUTION: Sketch the curve represented by the parametric equations. Then eliminate the parameter and write the corresponding rectangular equation whose graph represents the curve. Thanks so
Algebra.Com
Question 1080930: Sketch the curve represented by the parametric equations. Then eliminate the parameter and write the corresponding rectangular equation whose graph represents the curve. Thanks so much.
Sketch and Equations: https://s21.postimg.org/5tmtsylon/x_and_y_and_sketch.png
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
that straight line is extraneous and doesn't belong there.
in fact, that straight line is based on the equation of y = 1/2 * x -1
i did show it on the graph to confirm that the line was, in fact, based on that equation.
then i removed it.
your parametric equation is:
x = t^2 - 6
y = 1/2 * t - 1
in the desmos.com calculator, this is graphed as (t^2-6,1/2*t-1)
that calculator can be found here:
https://www.desmos.com/calculator
that means the x-coordinate on the graph is based on x = t^2 - 6 and the y-coordinate on the graph is based on y = 1/2 * t - 1
you do not see the value of t on the graph.
you only see the coordinate values of (x,y).
to convert from the parametric equation of x = t^2 - 6 and y = 1/2 * t - 1, you can do the f0llowing:
start with x = t^2 - 6
add 6 to both sides to get x + 6 = t^2
solve for t to get t = plus or minus sqrt(x+6)
in the equation of y = 1/2 * t - 1, replace t with plus or minus sqrt(x+6) to get:
when plus, y = 1/2 * t - 1 becomes y = 1/2 * sqrt(x+6) - 1
when minus, y = 1/2 * t - 1 becomes y = 1/2 * -sqrt(x+6) - 1 which becomes y = -1/2 * sqrt(x+6) - 1
your converted equation is therefore in two parts.
they are:
y = 1/2 * sqrt(x + 6) - 1 and y = -1/2 * sqrt(x + 6) - 1
the parametric equation and these translated equations are all shown in blue on the graph.
all 3 of them generate the same graph because the parametric equations is equivalent to the two y = ... equations.
when generating the parametric equations in the desmos.com calculator, you need to supply the limits for the value of t.
i made t go from -10 to 10 which was enough to show the graph in a similar manner as depicted on the image you provided.
the graph is shown below:
i also created a table of values based on the parametric equation that included the value of x and y when t = 0 and when t = -6 and when t = 6.
that table of values is shown below:
you can see from the table, that when t = 0, x = -6 and y = -1.
you can also see from the table, that when t = -6, x = 30 and y = -4.
you can also see from the table, that when t = 6, x = 30 and y = 2.
all these values can also be seen on the graph.
you see (-6,-1), (30,-4) and (30,2)
(-6,-1) is when t = 0
(30,-4) is when t = -6
(30,2)is when t = 6
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