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Chapter 11: Q.78 (page 1101)

In the following exercises, graph each equation by using properties.

x=-2(y-1)2+2

Short Answer

Expert verified

The graph of the equation is

Step by step solution

01

Step 1. Given Information

Given equation of the parabola isx=-2(y-1)2+2

The graph of the parabola is to be determined

02

Step 2. Equation of the parabola

The equation of a parabola in standard formx=a(y-k)2+h

The given equation isx=-2(y-1)2+2

Herea=-2,k=1,h=2

Since a is negative, the parabola opens to the left.

03

Step 3. Axis of symmetry

The axis of symmetry of the parabola x=a(y-k)2+hisy=k

Plugging the values:

y=1

Hence the axis of symmetry isy=1

04

Step 4. Vertex of a parabola

The vertex of the parabola x=a(y-k)2+his(h,k)

The vertex of the given parabola is(2,1)

05

Step 5. Intercepts

x-intercept is the point where the graph crosses the x-axis and y is 0.

y-intercept is the point where the graph crosses the y-axis and x is 0.

x-intercept:

Plugging y=0in the equation:

x=-2(y-1)2+2x=-2(0-1)2+2x=-2(-1)2+2x=-2(1)+2x=0

Hence the x-intercept is (0,0)


y-intercept:

Plugging x=0in the equation:

role="math" localid="1645432272221" x=-2(y-1)2+20=-2(y-1)2+2-2(y-1)2=-22(y-1)2=2(y-1)2=1y-1=卤1y-1=卤1y=卤1+1y=+1+1,y=-1+1y=2,y=0

Hence y-intercepts are (0,2),(0,0)

06

Step 6. Graphing the parabola

Using the properties, plotting the points and graphing the parabola:

07

Conclusion 

The graph of the parabola is

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