Graph f(x)=-x^2+2

Problem

ƒ(x)=−x2+2

Solution

1. Identify the type of function. This is a quadratic function in the form ƒ(x)=a*x2+b*x+c where a=−1 b=0 and c=2

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

3. Find the vertex. The xcoordinate is x=(−b)/(2*a)=0/(2*(−1))=0 Substituting x=0 into the function gives ƒ(0)=2 The vertex is (0,2)

4. Calculate the yintercept. Setting x=0 yields ƒ(0)=2 so the yintercept is (0,2)

5. Find the xintercepts. Set ƒ(x)=0 and solve −x2+2=0 which gives x2=2 so x=±√(,2)≈±1.41 The intercepts are (√(,2),0) and (−√(,2),0)

6. Plot additional points to refine the curve. For x=1 ƒ(1)=−(1)2+2=1 For x=2 ƒ(2)=−(2)2+2=−2 Due to symmetry across the yaxis (x=0, the points (−1,1) and (−2,−2) are also on the graph.

7. Sketch the downward-opening parabola passing through the vertex (0,2) and the calculated points.

Final Answer

ƒ(x)=−x2+2

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