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 The graph is a parabola that opens upward because a>0

2. Determine the vertex. Since b=0 the xcoordinate of the vertex is x=−b/(2*a)=0 Substituting x=0 into the function gives ƒ(0)=−2 The vertex is (0,−2)

3. Find the yintercept. Set x=0 to find the point where the graph crosses the yaxis.

ƒ(0)=0−2=−2

The yintercept is (0,−2)

4. Find the xintercepts. Set ƒ(x)=0 and solve for x

x2−2=0

x2=2

x=±√(,2)

The xintercepts are approximately (1.41,0) and (−1.41,0)

5. Plot additional points to define the shape.
   For x=1 ƒ(1)=1−2=−1 Point: (1,−1)
   For x=2 ƒ(2)=2−2=2 Point: (2,2)
   By symmetry across the yaxis (x=0, the points (−1,−1) and (−2,2) are also on the graph.

Final Answer

The graph of ƒ(x)=x2−2 is a parabola with vertex (0,−2) yintercept (0,−2) and xintercepts (±√(,2),0)

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