Question

How do you graph `f(x)=-(x-3)^2-2` and identify the x intercepts, vertex?

Answer 1

`f(x) = -(x-3)^2-2`

Please remember that
`f(x) = a(x-p)^2+p`,
means a = -ve, graph has max point and if a =+ve, graph has minimum point. The max or min point is at (p,q)

Therefore,
-ve sign infront of (x-3) means that it has a maximum point and it point / vertex is at (3,-2).

x intercept when f(x) =0.
therefore,

`0 = -(x-3)^2-2`
`(x-3)^2 =-2`
`(x-3) = sqrt(-2)`

This equation unable to determine, therefore there is no x-intercept or we can say that there is no real root for this equation.