( 1 ) Find by using a formula of (-b/2a, (4ac-b²)/4a)
[tex](\frac{-b}{2a},\frac{4ac-b^2}{4a})\\(\frac{-12}{2(-3)}),(\frac{4(-3)(-5)-(12)^2}{4(-3)})\\(\frac{-12}{-6}),(\frac{60-144}{-12})\\(2,\frac{-84}{-12})\\(2,7)[/tex]
The vertex is at (2,7)
( 2 ) Find by using a derivative (Calculus)
You can also find a vertex by using a calculus. Using a derivative
[tex]y'=-3x^2+12x-5\\y'=-6x+12\\\\[/tex]
Then substitute y' = 0 to get a vertex.
[tex]y'=-6x+12\\0=-6x+12\\-12=-6x\\2=x[/tex]
Then substitute x = 2 in the equation of y = -3x²+12x-5
[tex]y=-3x^2+12x-5\\y=-3(2)^2+12(2)-5\\y=-3(4)+24-5\\y=-12+24-5\\y=-17+24\\y=7[/tex]
x = 2 and y = 7, therefore. The vertex is (2,7)
