Graph y=-x^2+1

Problem

y=−x2+1

Solution

1. Identify the type of function. This is a quadratic function in the form y=a*x2+b*x+c where a=−1 b=0 and c=1 Since a<0 the parabola opens downward.

2. Find the vertex. The xcoordinate of the vertex is given by x=(−b)/(2*a) Substituting the values gives x=0/(2*(−1))=0 Substituting x=0 into the original equation gives y=−(0)2+1=1 The vertex is (0,1)

3. Determine the yintercept. Set x=0 in the equation.

y=−(0)2+1

y=1

The yintercept is (0,1) which is also the vertex.

4. Find the xintercepts. Set y=0 and solve for x

0=−x2+1

x2=1

x=±1

The xintercepts are (−1,0) and (1,0)

5. Plot additional points to define the shape. For x=2

y=−(2)2+1

y=−4+1

y=−3

For x=−2

y=−(−2)2+1

y=−4+1

y=−3

The points are (2,−3) and (−2,−3)

6. Sketch the curve. Draw a smooth, downward-opening parabola passing through the vertex (0,1) the xintercepts (−1,0) and (1,0) and the points (2,−3) and (−2,−3)

Final Answer

y=−x2+1

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