Graph y=-x^2-2x

Question

I don't know how to solve for x and y

Answer 1

Hello Jo.

In order to graph a parabola, all you have to do is make a function table and select various values of x and plug those values into the quadratic equation.  The only problem with this method is that you might come up  with something that does not look like a parabola, a U shape.  

 

So, in order to avoid, such problem , you need to start by finding the vertex of the parabola.  The formula for the vertex is

 (-b)/2a

 

The formula for a parabola is 

y= ax^2+bx+c.  

 

The problem that we need to graph is

y= x^2-2x

 

Therefore a=1

               b= (-2)

Then, the vertex will be

(2)/2= 1,  

 

Next, I am going to plug in 1 for x into our equation , y=x^2 -2x. 

y= (1)^2-2(1), y=-1 .  So, the vertex of this parabola is located as (1,-1).  

 

So, try to chose values of x's that are close to the vertext.  Let's choose for x: -2, -1, 0, 1, 2, 3 and plug these numbers into the equation        y= x^2-2x

Here we go:

x= -2, y=(-2)^2-2(-2)= 8.  So plot (-2, 8)

x= -1, y= (-1)^2-2(-1)=3.  So plot (-1, 3)

x=0, y=0.                          So plot (0,0). This also a x intercept.

x=1,    y=(1)^2-2(1)=-1    So plot (1,-1)  This is the vertex

x=2.    Y= (2)^2-2(2)=0.   So plot (2,0)   This is the other x intercept

x=3.     Y= (3)^2-2(3)=3.  So plot (3,3).  

 

If you plot all this points, you should get something that looks like a U.  This parabola is concave up.  

 

Now, to solve for y, make x=0, in this case (0,0) is both a y-intercept as well as one of the x intercepts. 

 

I hope this helps out.  I can't believe that it took me 40 minutes to type it.  But I am glad I could help. Thanks for posting your question.

 

D.Y. Taylor

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