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Chapter 11: Q. 342 (page 1160)

Identify the type of graph of each equation as a circle, parabola, ellipse, or hyperbola b) write the equation in standard form and c) graph the equation.

x=-y2+2y-4

Short Answer

Expert verified

The graph of the equation is parabola.

Step by step solution

01

Step 1. Given Information

Given that the equation isx=-y2+2y-4.

02

Step 2. Form of the graph 

When x and y is squared but not both called parabola. Hence, the graph is parabola.

03

Step 3. Standard form

From the equation, it is noted that,

a=-1,b=2,c=-4. The parabola open to the left when a=-1. So, the solution isx=-y2+2y-4.

04

Step 4. Find parameters

1) The horizontal parabola open to the left.

x=-y2+2y-4a=-1,b=2,c=-4

2) Axis of symmetry,

y=-b2ay=-22x-1y=1

The axis of symmetry is 1.

3) Vertex

x=-(1)2+2(1)-4x=-3

The vertex is (-3,1).

4) x-intercept

wheny=0,x=-(0)2+2(0)-4x=-4wheny=2,x=-(2)2+2(2)-4x=-4

The x-intercept are (-4,0) & (-4,2).

5) y-intercept

whenx=0,0=-y2+2y-4x=-b卤b2-4ac2ax=-1卤(-2)2-4脳1脳42脳1鈬�-1+-122,-1--122

y-intercept does not exist.

05

Step 5. Graph of the equation

On drawing the graph,

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