Question

How do you find the important points to graph \[f(x)={{x}^{2}}+2x\]?

Answer 1

Hint: In the given question, we are given an equation and we are asked to find the important points for the given equation and plot the equation. In order to find the important points of the given equation, we will put first \[x=0\] and then \[y=0\]. So, we will have the value of y – intercept and x – intercept. Then, we can also find the vertex of the given equation \[(h,k)\]. We now have the main points of the given equation, so now we can plot the required graph.

Complete step by step solution:
According to the given question, we are given an equation which we have to plot. So, we have to find the important points first.
The equation we have is,
\[f(x)={{x}^{2}}+2x\]---(1)
We will find the important points by finding the intercepts (both x and y). so, we will put first \[x=0\] and then \[y=0\], that is, we will be finding the y – intercept and the x – intercept respectively.
When we put \[x=0\], we get,
\[\Rightarrow f(0)=0+0=0\]
That is, the y-intercept is 0.
The coordinate we get is, \[\left( 0,0 \right)\].
When we put \[y=0\], the equation (1) can also be written as,
\[y={{x}^{2}}+2x\]----(2)
Substituting \[y=0\], we get,
\[0={{x}^{2}}+2x\]
We will now take x out as common and we get the expression as,
\[\Rightarrow x\left( x+2 \right)=0\]
Separating the components, we get,
\[x=0\] and \[x+2=0\] or \[x=-2\]
So, the coordinates we get are, \[\left( 0,0 \right)\] and \[\left( -2,0 \right)\].
From the equation itself we can understand that the given equation is the equation of a parabola.
The equation given to us \[y={{x}^{2}}+2x\] is similar to \[y=a{{x}^{2}}+bx+c\].
We can now also find the vertex \[(h,k)\], so we have,
\[h=-\dfrac{b}{2a}=-\dfrac{2}{2(1)}=-1\]
And \[k=c-\dfrac{{{b}^{2}}}{4a}=0-\dfrac{{{2}^{2}}}{4(1)}=-1\]
So, we have the vertex as \[(-1,-1)\].
Based on the above points, the graph we get is,

Note: The intercepts should be correctly known and accordingly only the graph will come out correctly. Y – intercept is the case when \[x=0\] and not \[y=0\]. Also, the equations \[y={{x}^{2}}\] give a parabola open upwards whereas \[y=-{{x}^{2}}\]gives a parabola which is open downwards.