Solveeit Logo

Question

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

How do you draw the graph y=x2+2xy={{x}^{2}}+2x?

Explanation

Solution

We find the xx-intercepts of the curve by putting y=0y=0 and we find yy-intercept by putting x=0x=0. We find the minimum value of the curve by making it a complete square. We take a minimum 5 points and join them with a free hand to get the plot.

Complete step-by-step solution:
We know that the graph of a function is the locus of points (x,y)\left( x,y \right) such that y=f(x)y=f\left( x \right) where x,yx,y are real numbers. We are given the following quadratic polynomial function
y=x2+2xy={{x}^{2}}+2x
So let us put y=y= and find the xx-intercept . We have

& 0={{x}^{2}}+2x \\\ & \Rightarrow x\left( x+2 \right)=0 \\\ & \Rightarrow x=0\text{ or }x=-2 \\\ \end{aligned}$$ So we find that $y=0$ at $x=0,-2$ .It means the curve cuts $x-$axis at $\left( -2,0 \right),\left( 0,0 \right)$.Let us put $x=0$ in the curve to have; $$y={{0}^{2}}+2\cdot 0=0$$ So there is no $y-$intercept of the curve. We know that all quadratic functions of the type $y=a{{x}^{2}}+bx+c$ have minimum values but not maximum. So let us complete square in $x$ to have; $$\begin{aligned} & y={{x}^{2}}+2x \\\ & \Rightarrow y={{x}^{2}}+2x+1-1 \\\ & \Rightarrow y={{\left( x+1 \right)}^{2}}-1 \\\ \end{aligned}$$ Since square is always non-negative we have ${{\left( x+1 \right)}^{2}}\ge 0$, then we have $$\begin{aligned} & \Rightarrow y={{\left( x+1 \right)}^{2}}-1\ge -1 \\\ & \Rightarrow y\ge -1 \\\ \end{aligned}$$ So the minimum value of $y=-1$ and the minimum value occurs when ${{\left( x+1 \right)}^{2}}=0$ or $x=-1$. We have already three points for the curve $\left( -2,0 \right),\left( -1,1 \right),\left( 0,0 \right)$. We find $y$ for two more points at $x=-3,x=1$ to have $$\begin{aligned} & y\left( -3 \right)={{\left( -3 \right)}^{2}}+2\left( -3 \right)=3 \\\ & y\left( 1 \right)={{1}^{2}}+2\cdot 1=3 \\\ \end{aligned}$$ So we draw the table for $x$ and $y$. $x$| $y$ ---|--- $-3$| 3 $-2$| 0 $-1$| $-1$ 0| 0 1| 1 We plot the above points and join them to have the graph as ![](https://www.vedantu.com/question-sets/6a4429de-2a62-49b5-b6b3-ae67b29bbbca4063343069686637644.png) **Note:** We note that the obtained graph is the graph of upward parabola whose general equation is given by $y=a{{x}^{2}}+bx+c$ with the condition $a>0$ whose vertex here is $\left( -1,-1 \right)$.We can directly find the minimum value of $y={{x}^{2}}+2x$ by finding $y$ at $x=\dfrac{-b}{2a}$. If $a<0$ we the equation $y=a{{x}^{2}}+bx+c$ represents a downward parabola. We also note that the obtained curve is symmetric about the line $x=-1$.