Graph f(x)=x^2-5

Problem

ƒ(x)=x2−5

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=−5 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)=−5 The vertex is (0,−5)

3. Find the yintercept. Set x=0 to find ƒ(0)=−5 The yintercept is (0,−5) which is also the vertex.

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

x2−5=0

x2=5

x=±√(,5)

The xintercepts are approximately (−2.24,0) and (2.24,0)

5. Plot additional points to define the shape. For x=1 ƒ(1)=1−5=−4 For x=2 ƒ(2)=2−5=−1 Due to symmetry across the yaxis (x=0, the points (−1,−4) and (−2,−1) are also on the graph.

6. Sketch the curve. Draw a smooth, U-shaped curve passing through the vertex (0,−5) the intercepts, and the additional points.

Final Answer

ƒ(x)=x2−5

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