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+c which represents a parabola.

2. Determine the vertex. Since there is no x term (b=0, the xcoordinate of the vertex is 0 Substituting x=0 into the function gives ƒ(0)=5 so the vertex is at (0,5)

3. Analyze the direction of the opening. The coefficient of x2 is a=1 Since a>0 the parabola opens upward.

4. Find additional points to plot. Choose x values around the vertex:

   • If x=1 ƒ(1)=1+5=6 Point: (1,6)

   • If x=−1 ƒ*(−1)=(−1)2+5=6 Point: (−1,6)

   • If x=2 ƒ(2)=2+5=9 Point: (2,9)

   • If x=−2 ƒ*(−2)=(−2)2+5=9 Point: (−2,9)

5. Sketch the graph by plotting the vertex (0,5) and the calculated points, then drawing a smooth U-shaped curve through them.

Final Answer

ƒ(x)=x2+5* is a parabola with vertex at *(0,5)* opening upward.

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