Question

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

Answer 1

Hint: We equate the given equation of parabolic curve with the general equation of ${{\left( x-\alpha \right)}^{2}}=4a\left( y-\beta \right)$. 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 $y={{x}^{2}}+2x+1$ is a parabolic curve. We equate it with the general equation of parabola ${{\left( x-\alpha \right)}^{2}}=4a\left( y-\beta \right)$.
For the general equation $\left( \alpha ,\beta \right)$ is the vertex. 4a is the length of the latus rectum. The coordinate of the focus is $\left( \alpha ,\beta +a \right)$.
Now we convert the given equation $y={{x}^{2}}+2x+1$ according to the general equation to find the value of the vertex.
We get
$\begin{align}
  & y={{x}^{2}}+2x+1 \\
 & \Rightarrow {{\left( x+1 \right)}^{2}}=\left( y+0 \right) \\
\end{align}$
This gives the vertex as $\left( -1,0 \right)$. The length of the latus rectum is $4a=1$ which gives $a=\dfrac{1}{4}$.
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 $x=0$ in the equation $y={{x}^{2}}+2x+1$, we get
$y={{0}^{2}}+2\times 0+1=1$
The intercept is the point $\left( 0,1 \right)$. 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 $y=0$ in the equation $y={{x}^{2}}+2x+1$, we get
\[\begin{align}
  & y={{x}^{2}}+2x+1 \\
 & \Rightarrow {{\left( x+1 \right)}^{2}}=0 \\
 & \Rightarrow x=-1 \\
\end{align}\]
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.