`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` .
