Answer :
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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The vertex is (-2, -1). The axis of symmetry is x = -2. The minimum value is -1. The maximum value is infinity.
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
Answer and Explanation :
Given : Function [tex]f(x)=\frac{x^2}{2}+2x+1[/tex]
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, [tex]y=ax^2+bx+c[/tex]
On comparing, [tex]a=\frac{1}{2}[/tex] , 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,
[tex]h=-\frac{b}{2a}[/tex]
[tex]h=-\frac{2}{2(\frac{1}{2})}[/tex]
[tex]h=-\frac{2}{1}[/tex]
[tex]h=-2[/tex]
For k, The y-coordinate of the vertex is given by,
[tex]k=f(h)[/tex]
[tex]k=\frac{h^2}{2}+2h+1[/tex]
[tex]k=\frac{(-2)^2}{2}+2(-2)+1[/tex]
[tex]k=2-4+1[/tex]
[tex]k=-1[/tex]
The vertex of the function is (h,k)=(-2,-1)
The x-coordinate of the vertex i.e. [tex]x=-\frac{b}{2a}[/tex] is the axis of symmetry,
So, [tex]x=-\frac{b}{2a}=-2[/tex] (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.
[tex]a=\frac{1}{2} >0[/tex] 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.
