Respuesta :
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!!!
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:
Vertex = (-1, 0)
The axis of symmetry is x = -1
Minimum point is (-1, 0)
Range is y ≥ 0
In the interval notation [0, ∞)
Step-by-step 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.