Graph -x^2+1

Problem

−x2+1

Solution

1. Identify the type of function. The expression ƒ(x)=−x2+1 is a quadratic function in the form ƒ(x)=a*x2+b*x+c where a=−1 b=0 and c=1

2. Determine the shape and orientation. Since the leading coefficient a=−1 is negative, the parabola opens downward.

3. Find the vertex. The xcoordinate of the vertex is given by x=(−b)/(2*a)

x=0/(2*(−1))=0

Substituting x=0 into the function gives the ycoordinate.

y=−(0)2+1=1

The vertex is at (0,1)

4. Find the xintercepts by setting ƒ(x)=0

−x2+1=0

x2=1

x=±1

The xintercepts are at (1,0) and (−1,0)

5. Find the yintercept by setting x=0

y=−(0)2+1=1

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

6. Plot the points and draw a smooth curve. The graph is a downward-opening parabola with its highest point at (0,1) and crossing the xaxis at x=−1 and x=1

Final Answer

The graph of −x2+1 is a downward-opening parabola with vertex (0,1) and xintercepts (−1,0) and (1,0)

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