Graph y=-2x^2+2

Problem

y=−2*x2+2

Solution

1. Identify the function type and its basic shape. This is a quadratic function in the form y=a*x2+b*x+c which represents a parabola. Since a=−2 the parabola opens downward.

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

3. Determine the y-intercept by setting x=0 As calculated for the vertex, the yintercept is (0,2)

4. Find the x-intercepts by setting y=0 and solving for x

0=−2*x2+2

−2=−2*x2

1=x2

x=±1

The xintercepts are (−1,0) and (1,0)

5. Plot additional points to refine the curve. For x=2

y=−2*(2)2+2

y=−2*(4)+2

y=−6

The point is (2,−6) By symmetry, the point (−2,−6) is also on the graph.

6. Sketch the graph by drawing a smooth, downward-opening curve through the points (−2,−6) (−1,0) (0,2) (1,0) and (2,−6)

Final Answer

y=−2*x2+2

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