Question
Question: How do you graph \(y={{x}^{2}}+2x+1\)?...
How do you graph y=x2+2x+1?
Solution
We equate the given equation of parabolic curve with the general equation of (x−α)2=4a(y−β). We find the number of x intercepts and the value of the y intercept. We also find the coordinates of the focus to place the curve in the graph.
Complete step-by-step answer:
The given equation y=x2+2x+1 is a parabolic curve. We equate it with the general equation of parabola (x−α)2=4a(y−β).
For the general equation (α,β) is the vertex. 4a is the length of the latus rectum. The coordinate of the focus is (α,β+a).
Now we convert the given equation y=x2+2x+1 according to the general equation to find the value of the vertex.
We get
y=x2+2x+1⇒(x+1)2=(y+0)
This gives the vertex as (−1,0). The length of the latus rectum is 4a=1 which gives a=41.
We have to find the possible number of x intercepts and the value of the y intercept. The curve cuts the X and Y axis at certain points and those are the intercepts.
We first find the Y-axis intercepts. In that case for the Y-axis, we have to take the coordinate values of x as 0. Putting the value of x=0 in the equation y=x2+2x+1, we get
y=02+2×0+1=1
The intercept is the point (0,1). The vertex is the intercept and it’s the only intercept on the Y-axis.
We first find the X-axis intercepts. In that case for X-axis, we have to take the coordinate values of y as 0. Putting the value of y=0 in the equation y=x2+2x+1, we get