`x = 5-4t`

`y=2+5t`

To graph a parametric equation, assign values to t. Since there is no given interval for t, let's consider the values from t=-3 to t=3.

t=-3

`x=5-4(-3) = 17`

`y= 2+5(-3) = -13`

t=-2

`x=5-4(-2)=13`

`y=2+5(-2)=-8`

t=-1

`x=5-4(-1)=9`

`y=2+5(-1)=-3`

t=0

`x=5-4(0)=5`

`y=2+5(0)=2`

t=1

`x=5-4(1)=1`

`y=2+5(1)=7`

t=2

...

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`x = 5-4t`

`y=2+5t`

To graph a parametric equation, assign values to t. Since there is no given interval for t, let's consider the values from t=-3 to t=3.

t=-3

`x=5-4(-3) = 17`

`y= 2+5(-3) = -13`

t=-2

`x=5-4(-2)=13`

`y=2+5(-2)=-8`

t=-1

`x=5-4(-1)=9`

`y=2+5(-1)=-3`

t=0

`x=5-4(0)=5`

`y=2+5(0)=2`

t=1

`x=5-4(1)=1`

`y=2+5(1)=7`

t=2

`x=5-4(2)=-3`

`y=2+5(2)=12`

t=3

`x=5-4(3)=-7`

`y=2+5(3)=17`

And, plot the points (x,y) in the xy-plane.

(Please see attachment for the orientation of the curve.)

Take note that the graph of a parametric equation has direction. For this equation, as the value of t increases, the points (x,y) are going to upward to the left.

To convert a parametric equation to rectangular form, isolate the t in one of the equation. Let's consider the equation for x.

`x= 5-4t`

`x-5=-4t`

`-(x-5)/4=t`

Then, plug-in this to the other equation.

`y=2+5(t)`

`y=2+5(-(x-5)/4)`

`y=2-(5(x-5))/4`

`y=2-(5(x-5))/4`

`y=2-(5x - 25)/4`

`y=8/4-(5x-25)/4`

`y=(33 - 5x)/4`

`y=-(5x)/4 + 33/4`

**Therefore, the rectangular form of the given parametric equation is `y=-(5x)/4 + 33/4` .**