Graph x=y^2-1

Problem

x=y2−1

Solution

1. Identify the type of curve. The equation x=y2−1 is a parabola that opens horizontally because the y term is squared and the x term is linear. Since the coefficient of y2 is positive, the parabola opens to the right.

2. Find the vertex. The standard form for a horizontal parabola is x=a*(y−k)2+h where (h,k) is the vertex. Rewriting the equation as x=1*(y−0)2−1 we identify the vertex at (−1,0)

3. Determine the x-intercept. Set y=0 in the equation.

x=(0)2−1

x=−1

The x-intercept is (−1,0) which is also the vertex.

4. Determine the y-intercepts. Set x=0 and solve for y

0=y2−1

y2=1

y=±1

The y-intercepts are (0,1) and (0,−1)

5. Plot additional points to define the shape. For example, if y=2 or y=−2

x=(±2)2−1

x=4−1

x=3

The points (3,2) and (3,−2) are on the graph.

6. Sketch the curve. Draw a smooth curve passing through the vertex (−1,0) the y-intercepts (0,1) and (0,−1) and the points (3,2) and (3,−2) opening to the right.

Final Answer

To graph x=y2−1 plot the vertex at (−1,0) the y-intercepts at (0,1) and (0,−1) and draw a parabola opening to the right.

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