Question

In the following exercise graph each equation by using its properties:

$x=2y^2+4y+6$

Short answer

The graph of the parabola can be given as

Step-by-step solution

Given information

We are given a equation$x=2y^2+4y+6$

find whether the parabola opens Right or Left

As the coefficient of square term is positive the parabola opens to the Right

Find the axis of symmetry

To find axis of symmetry we substitute$$\begin{gathered} y=-\frac{b}{2a} \\ y=-\frac{4}{4} \\ y=-1 \end{gathered}$$

Find the vertex of parabola

To find the vertex put $y=-1$we get

$$\begin{gathered} x=2y^2+4y+6 \\ x=2-4+6 \\ x=4 \end{gathered}$$

The vertex of parabola is$(4,-1)$

Find the x-intercept

Put $y=0$

we get,

$$\begin{gathered} x=2y^2+4y+6 \\ x=0+0+6 \\ x=6 \end{gathered}$$

The coordinates are$(6,0)$

Find the y-intercept

Put $x=0$

$$\begin{gathered} 2y^2+4y+6=0 \\ {y}^2+2y+3=0 \end{gathered}$$

On solving further we conclude that the roots of the above equation are imaginary

Hence the y-intercept is absent

Plot the graph of parabola 

The graph of parabola can be given as

Conclusion

The graph of parabola can be given as