Graph y=-x^2+3

Problem

y=−x2+3

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=3

2. Determine the shape and orientation. Since 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

4. Calculate the ycoordinate of the vertex by substituting x=0 into the equation.

y=−(0)2+3=3

The vertex is at (0,3)

5. Find the yintercept. Set x=0

y=−(0)2+3=3

The yintercept is (0,3)

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

0=−x2+3

x2=3

x=±√(,3)≈±1.732

The xintercepts are (√(,3),0) and (−√(,3),0)

7. Plot additional points to refine the curve. For x=1

y=−(1)2+3=2

For x=2

y=−(2)2+3=−1

8. Sketch the graph by drawing a smooth, downward-opening curve through the points (0,3) (1,2) (−1,2) (2,−1) and (−2,−1)

Final Answer

y=−x2+3

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