**$(a) \space x=t^4 -t+1 , y= t^2$**

**$(b) \space x=t^2 -2t , y=\sqrt t$**

**$(c) \space\ x=\sin2t ,y=\sin ( t +\sin 2t)$**

**$(d) \space x=\cos5t ,y=\sin 2t$**

**$(e) \space x=t+\sin4t ,y= t^2 +\cos3t$**

**$(f) \space x=\dfrac{\sin2t }{4+t^2} ,y=\dfrac{\cos2t} {4+t^2}$**

**Graph I**

**Graph II**

**Graph III**

**Graph IV**

**Graph V**

**Graph VI**

In this question, we have to match the given **functions** with the given **graphs** labeled from **I to VI**. For this, we have to recall our fundamental knowledge of **Calculus **for the** most suitable match **of the** functions **with the given** graphs**.

This question uses the basic concepts of **Calculus** and **Linear Algebra** by **matching** the functions to the **best** graphs.

**Expert Answer**

**$(a) \space x=t^4 -t+1 , y= t^2$:**

For the given **parametric equation**, suppose the value of $t$ is equal to **zero**, then we have the function equal to:

\[x=(0)^4 -0+1\ ,\ y= (0)^2\]

\[ x= 1 , y= 0\]

When the value of $t$ is **zero** then $x=1$ and $y=0$, there is no other graph with starts at $x=1$. So, for this equation, the **best graph is labeled **$V$.

**Graph V**

**$(b) \space x= t^2 -2t , y= \sqrt t$**

For the given **parametric equation**, suppose the value of $t$ is equal to **zero**, then we have the function equal to:

\[x=(0)^2 -2t\ ,\ y= \sqrt (0)\]

\[x= 0 , y= 0\]

When the value of $t$ is **zero,** then $x=0$ and $y=0$. There is no other graph with starts at $x=0$ and both the coordinate values go to **infinity**, so for this equation, the **best graph is labeled** $I$.

**Graph I**

**$(c) \space\ x= \sin2t ,y= \sin ( t +\sin 2t)$**

For the given **parametric equation**, when the value of $t$ is **zero,** then $x=0$ and $y=0$. There is no other graph that has the value of $(0,1)$, which is at $t=\dfrac{\pi}{2}$. So, for this equation, the **best graph is labeled **$II$.

**Graph II**

**$(d) \space x= \cos5t ,y= \sin 2t $**

For the given **parametric equation**, when the value of $t$ is **zero**, then $x=1$ and $y=0$. There is no other graph that has the value of $(0,1)$ which is at $t=0$. So, for this equation, the **best graph is labeled **$IV$.

**Graph IV**

**$(e) \space x= t+ \sin 4t ,y= t^2 +\cos3t $**

For the given **parametric equation**, the value of **both coordinates **$x$ and $y$ goes to** infinity.** There is no other graph that also shows the **oscillatory behavior**. So, the **best graph is labeled **$VI$.

**Graph VI**

**$(f)\ x= \dfrac{\sin 2 t }{4 + t^2} ,y= \dfrac { \cos2 t} {4+ t^2 }$**

For the given **parametric equation**, the value of both **coordinates **$x$ and $y$ cannot be $(0,0)$ but with the **oscillatory behavior**. So the **best graph is labeled **$III$.

**Graph III**

**Numerical Result**

By assuming the values of $x$ and $y$, functions are matched with the best** graphs**.

**Example**

Draw the **graph** for **function** **$(x,y)=(\sin t-7t,\ \sin\ 2t)$.**

Put $t=0$ , $t=\dfrac{\pi}{2}$

The** graph** for the **given function** is as follows:

**Figure I**

*Images/Mathematical drawings are created with Geogebra.*