Question

Identify the​ vertex, the axis of​ symmetry, the maximum or minimum​ value, and the range of the parabola.

yequalsxsquaredplus2xplus1
The vertex is
nothing. ​(Type an ordered​ pair.)

Answer 1

Y = x^2 + 2x + 1
Convert to vertex form by completing the square
(2/2)^2 = 1
(X^2+2x+1) -1 +1
Y = (X + 1)^2
The vertex is (1,0) because we have no + q, and we have 1 as our p, and since I don’t know any of the points in order to figure out the vertical stretch, this is the final answer.

Hope this helped!!!

Answer 2

Vertex = (-1, 0)

The axis of symmetry is x = -1

Minimum point is (-1, 0)

Range is y ≥ 0

In the interval notation [0, ∞)

Explanation:

The given function is y = x^2 +2x + 1.

Write the given equation in the form of y = (x -h)^2 + k.

Here (h,k) is the vertex.

y = x^2 + 2x + 1

y = (x + 1)^2

Vertex = (-1, 0)

Axis of symmetry

The axis of symmetry splits the parabola in 2 equal parts.

In this equation, the axis of symmetry is x = -1

Maximum/minimum value

The parabola goes maximum when there is negative sign, it goes minimum when there is positive sign.

Here we have positive sign in front of (x + 1)^2, therefore, it is minimum at

y= 0

When y =0, x = -1

Therefore, minimum point is (-1, 0)

Range

Y- axis represents the range.

Here the minimum is 0, the maximum goes upto infinity.

Range is y ≥ 0

In the interval notation [0, ∞)

Herewith I have attached the graph of the function.