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Question: How do you graph \(y={{x}^{2}}+2x+1\)?...

How do you graph y=x2+2x+1y={{x}^{2}}+2x+1?

Explanation

Solution

We equate the given equation of parabolic curve with the general equation of (xα)2=4a(yβ){{\left( x-\alpha \right)}^{2}}=4a\left( y-\beta \right). 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+1y={{x}^{2}}+2x+1 is a parabolic curve. We equate it with the general equation of parabola (xα)2=4a(yβ){{\left( x-\alpha \right)}^{2}}=4a\left( y-\beta \right).
For the general equation (α,β)\left( \alpha ,\beta \right) is the vertex. 4a is the length of the latus rectum. The coordinate of the focus is (α,β+a)\left( \alpha ,\beta +a \right).
Now we convert the given equation y=x2+2x+1y={{x}^{2}}+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) \begin{aligned} & y={{x}^{2}}+2x+1 \\\ & \Rightarrow {{\left( x+1 \right)}^{2}}=\left( y+0 \right) \\\ \end{aligned}
This gives the vertex as (1,0)\left( -1,0 \right). The length of the latus rectum is 4a=14a=1 which gives a=14a=\dfrac{1}{4}.
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=0x=0 in the equation y=x2+2x+1y={{x}^{2}}+2x+1, we get
y=02+2×0+1=1y={{0}^{2}}+2\times 0+1=1
The intercept is the point (0,1)\left( 0,1 \right). 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=0y=0 in the equation y=x2+2x+1y={{x}^{2}}+2x+1, we get

& y={{x}^{2}}+2x+1 \\\ & \Rightarrow {{\left( x+1 \right)}^{2}}=0 \\\ & \Rightarrow x=-1 \\\ \end{aligned}$$ The intercept points are $\left( -1,0 \right)$. There is only one intercept on X-axis. **Note:** The minimum point of the function $y={{x}^{2}}+2x+1$ is $y=0$. The graph is bounded at that point. But on the other side the curve is open and not bounded. the general case of a parabolic curve is to be bounded at one side to mark the vertex.