f(x)-(x+3)^2-4

Question

Find the vertex then graph using a table. 

f(x)-(x+3)^2-4

Answer 1

Hello Macz, 

I think you forgot the equal sign.  Ok, this equation is already writing in the Vertex Form which is

 

y= a( x-h)^2+k, where (h, k) represent the coordinate of the vertex.  

y= (x+3)^2-4    has its vertex at (-3, -4)

 

When I teach parabolas, some of my students plot the points correctly, but they don't come up with the U shape.  Yes, theoretically, you can choose any value for x.  But if you want the graph to look like a parabola, you have to choose the x values carefully.  You do this by looking at the x value of the vertex.  In this case, it is -3.  So, I am going to choose a couple x values lower than -3 and a couple of x values higher than -3. So, I will use -5,-4,-3,-2,-1 as my values.

 

Now it comes the tedious work.  My students love this since they don't have to think !!!! But, I digress.

 

y= (x+3)^2-4

y=(-5+3)^2-4 = (-2)^2-4= 4-4= 0.       Plot (-5, 0)

y=(-4+3)^2-4=  (-1)^2-4= 1-4=-3.       Plot (-4, -3)

y= (-3+3)^2-4= (0)^2-4=  0-4=-4.        Plot (-3, -4). Vertex

y= (-2+3)^2-4=  (1)^2-4= 1-4=-3.        Plot (-2,-3)

y= (-1+3)^2-4=  (2)^2-4= 4-4=0.          Plot (-1, 0)

 

When you plot these points, you should get a symmetrical parabola. A nice U- shape.  If you want more points try 0,1,2,-6,-7,-8 for x.  

 

Well. It took me long, but it is worth it.  I hope this helps out. 

 

D.Y. T.

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