Graph f(x)=-x^2+4

Problem

ƒ(x)=−x2+4

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=4

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) Since b=0 the xcoordinate is 0 Substituting x=0 into the function gives ƒ(0)=4 so the vertex is (0,4)

4. Find the y-intercept by evaluating ƒ(0) The yintercept is (0,4) which is also the vertex.

5. Find the x-intercepts by setting ƒ(x)=0 Solving −x2+4=0 gives x2=4 resulting in x=2 and x=−2 The intercepts are (2,0) and (−2,0)

6. Plot the points and draw a smooth downward-opening curve through (−2,0) (0,4) and (2,0)

Final Answer

ƒ(x)=−x2+4

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