Graph y=x^2+2x+1

Problem

y=x2+2*x+1

Solution

1. Identify the type of function. This is a quadratic function in the standard form y=a*x2+b*x+c where a=1 b=2 and c=1 Since a>0 the parabola opens upward.

2. Factor the quadratic expression to find the xintercepts. The expression x2+2*x+1 is a perfect square trinomial.

y=(x+1)2

3. Find the xintercept by setting y=0

0=(x+1)2

x=−1

The graph touches the xaxis at the point (−1,0)

4. Determine the vertex of the parabola. Since there is only one xintercept, the vertex is located at that point.

Vertex=(−1,0)

5. Find the yintercept by setting x=0

y=0+2*(0)+1

y=1

The yintercept is at the point (0,1)

6. Identify additional points to define the shape. Using the axis of symmetry x=−1 find the symmetric point to the yintercept.

x=−2⇒y=(−2)2+2*(−2)+1=1

The point is (−2,1)

Final Answer

y=(x+1)2

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