For the function below, find the vertex, axis of symmetry, maximum or minimum value, and the graph of the function.
f(x)=(x^2/2) + 2x +1 ...?
f(x) = x^2/2 + 2x + 1f(x) = 1/2 (x^2 + 4x) + 1f(x) = 1/2 (x^2 + 4x + 4) - 1/2 (4) + 1f(x) = 1/2 (x + 2)^2 - 1
The vertex is (-2, -1). The axis of symmetry is x = -2. The minimum value is -1. The maximum value is infinity.
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Answer and Explanation :
Given : Function \(f(x)=\frac{x^2}{2}+2x+1\)
To find : The vertex, axis of symmetry, maximum or minimum value, and the graph of the function.
Solution :
The quadratic function is in the form, \(y=ax^2+bx+c\)
On comparing, \(a=\frac{1}{2}\) , b=2 and c=1
The vertex of the graph is denote by (h,k) and the formula to find the vertex is
For h, The x-coordinate of the vertex is given by,
\(h=-\frac{b}{2a}\)
\(h=-\frac{2}{2(\frac{1}{2})}\)
\(h=-\frac{2}{1}\)
\(h=-2\)
For k, The y-coordinate of the vertex is given by,
\(k=f(h)\)
\(k=\frac{h^2}{2}+2h+1\)
\(k=\frac{(-2)^2}{2}+2(-2)+1\)
\(k=2-4+1\)
\(k=-1\)
The vertex of the function is (h,k)=(-2,-1)
The x-coordinate of the vertex i.e. \(x=-\frac{b}{2a}\) is the axis of symmetry,
So, \(x=-\frac{b}{2a}=-2\) (solved above)
So, The axis of symmetry is x=-2.
The maximum or minimum point is determine by,
If a > 0 (positive), then the parabola opens upward and the graph has a minimum at its vertex.
\(a=\frac{1}{2} >0\) so, the parabola opens upward and the graph has a minimum at its vertex.
The Minimum value is given at (-2,-1)
Now, We plot the graph of the function
At different points,
x y
-4 1
-2 -1
0 1
Refer the attached figure below.
