Q. 19 - Calculus

Question

Match the two graphs shown here to the equations $\left(x+1\right)\left(y^2+y-1\right)=1$ and $x{y}^2+y=1$Explain your choices.

 

Step-by-Step Solution

Answer

The graph of $\left(x+1\right)\left(y^2+y-1\right)=1$

The graph of $x{y}^2+y=1$

Step 1. Given information

The functions are:

First function: $\left(x+1\right)\left(y^2+y-1\right)=1$

Second function: $x{y}^2+y=1$

The graph;

Step 2. Find x - intercepts of both functions.

Substitute $x=0$ in both functions and solve for y.

$$\begin{gathered} \left(x+1\right)\left(y^2+y-1\right)=1 \\ \left(0+1\right)(y^2+y-1)=1 \\ {y}^2+y-1=1 \\ {y}^2+y-2=0 \\ {y}^2+2y-y-2=0 \\ y(y+2)-1(y+2)=0 \\ (y-1)(y+2)=0 \\ y-1=0 \\ y=1 \\ y+2=0 \\ y=-2 \end{gathered}$$

So $y-$intercepts of the first function are:$\left(0,-2\right)$ and $\left(0,1\right)$.

Similarly in second function:

$$\begin{gathered} 0.y^2+y+1=0 \\ y=-1 \end{gathered}$$

So $y-$intercept of the second function is $\left(0,-1\right)$

Step 3. Match the graphs with their function.

Since $\left(x+1\right)\left(y^2+y-1\right)=1$, has $y-$intercept are $(0,-2)$ and $(0,1)$,so the graph will cross at these points.

So the graph is:

The $y-$intercept of $x{y}^2+y=1$ is $\left(0,1\right)$

So the graph of the function crosses $y-$axis at this point.