I need vertex,
line of Sym x=
Max and Min
is the F(0)=5 a min or a max?
Much appriciated
vertex is (0,5)
line of sym is x=0
take derative of function and equate to 0.
y'= -2x
0= -2x
so x = 0 this is either a min or max. Now you can use the 2nd derative test or plug in values for the 1st derative and see where it increasing or decreasing. For example -1<0<1
y' = -2(-1)= 2
y' = -2(1) = -2
so when x=0 is a max
Hope that helps
You don't need to differentiate.
We are looking for the maximum value of $\displaystyle 5 - x^2$. It's obvious that $\displaystyle x^2$ is always positive or zero. So, the minimum of $\displaystyle x^2$ is 0. Therefore $\displaystyle 5-x^2$ is maximum at $\displaystyle x=0$.
We can also find it using differentials. The local extrema of a function are at the roots of $\displaystyle y' = 0$. Don't forget to check for global extrema too.
$\displaystyle y' = -2x = 0$
$\displaystyle x=0$
So we have a local extremum at $\displaystyle x=0$.
We can decide whether it's maximum or minimum using the second derivative test.
$\displaystyle y''(0) = -2 < 0$. It's a maximum.
Don't forget to check for global maximum too. The reason for doing this is, the function may have a maximum at infinity.
$\displaystyle \lim_{x\to\infty} 5-x^2 = -\infty$
$\displaystyle \lim_{x\to-\infty} 5-x^2 = -\infty$
Ok, now we're convinced that the function has a maximum only at $\displaystyle x=0$.