Question

In the following exercise a) write the equation in standard form b) use properties of standard form to plot the graph:$x=-2y^2+4y$

Short answer

The equation in standard form can be given as$x=-2{(y-1)}^2+2$

The graph of the parabola can be given as

Step-by-step solution

Given information

We are given an equation$x=-2y^2+4y$

Write the equation in standard form 

We have,

$$\begin{gathered} x=-2y^2+4y \\ x=-2(y^2-2y) \\ x=-2(y^2-2y+1)+2 \\ x=-{(y-1)}^2+2 \end{gathered}$$

On comparing with standard equation we get$a=-1,k=1,h=2$

Find the axis of symmetry and vertex of parabola

The axis of symmetry can be given as $y=k$

we have $y=1$

And the vertex can be given asrole="math" localid="1645511144589" $(k,h)=(2,1)$

Find x-intercept

Put $y=0$

we get

$x=0$

Find y-intercept

Put $x=0$

we get

$$\begin{gathered} -2y^2+4y=0 \\ {y}^2-2y=0 \\ y(y-2)=0 \\ y=0ory=2 \end{gathered}$$

The coordinates are$(0,0),(0,2)$

Plot the graph

The graph can be given as

Conclusion

The equation in standard form can be given as$x=-2{(y-1)}^2+2$

The graph of the parabola can be given as