Graph f(x)=-x^2+3

Problem

ƒ(x)=−x2+3

Solution

1. Identify the function type, which is a quadratic function in the form ƒ(x)=a*x2+b*x+c Here, a=−1 b=0 and c=3

2. Determine the concavity by looking at the leading coefficient a Since a=−1 is negative, the parabola opens downward.

3. Find the vertex using the formula x=(−b)/(2*a)

x=0/(2*(−1))=0

Substitute x=0 into the function to find the ycoordinate.

ƒ(0)=−(0)2+3=3

The vertex is (0,3)

4. Find the y-intercept by evaluating ƒ(0)

ƒ(0)=3

The yintercept is (0,3)

5. Find the x-intercepts by setting ƒ(x)=0

−x2+3=0

x2=3

x=±√(,3)≈±1.732

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

6. Plot additional points to refine the shape.
   For x=1

ƒ(1)=−(1)2+3=2

For x=2

ƒ(2)=−(2)2+3=−1

Due to symmetry across the yaxis (x=0, the points (−1,2) and (−2,−1) are also on the graph.

Final Answer

The graph of ƒ(x)=−x2+3 is a downward-opening parabola with vertex (0,3) yintercept (0,3) and xintercepts (±√(,3),0)

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