Question

For the function below, find the vertex, axis of symmetry, maximum or minimum value, and the graph of the function.

Answer 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 2

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.