what is the vertex of f(x)=x^2+2

what is the vertex of f(x)=x^2+2

if possible, please show steps, thank you.

Question

what is the vertex of f(x)=x^2+2

what is the vertex of f(x)=x^2+2 

if possible, please show steps, thank you.

anonymous · Answer

There are a number of ways:
A) Plot the  graph of the equation on the calculator  and find the coordinates  at the trough of the graph.

|dw:1332891720924:dw| You can use calculus by differentiating the equation and setting the derivative to 0, then solving for x.  Once the value of x is found plug that value in the original equation to get the y value:
i.e.
$$y=x ^{2}  +2$$ y' = 2x
Set y' =0
=> y'=2x=0
=> x=0
f(0) =2
Therefore the vertex is at (0, 2)

C) Use the  b/2a. This gives the x value of the vertex.
Plug the x value in the original equation to get the y value.

i.e 
for  an equation:

$$y= ax ^{2} + bx +c  $$set y =0
$$ax ^{2}  + bx + C  =0$$ The x value of the  vertex is  b/2a

Your equation:
$$y=x ^{2}  +2$$ => a=1, b=0, c=2
i.e. it is equivalent to  $$y=x ^{2} +0x  + 2$$
X value of the vertex = 0/2  = 0
substitute the x value back into the original equation to get y: 
thus:  $$y=0^{2} + 2  =2$$ The vertex is therefore  (0,2).