Question

2.) Find the vertex only for the following function:
y = -3x^2 + 12x - 5

Answer 1

(2, 7)

Explanation:

hope this helps :)

Answer 2

( 1 ) Find by using a formula of (-b/2a, (4ac-b²)/4a)

`((-b)/(2a),(4ac-b^2)/(4a))\\((-12)/(2(-3))),((4(-3)(-5)-(12)^2)/(4(-3)))\\((-12)/(-6)),((60-144)/(-12))\\(2,(-84)/(-12))\\(2,7)`

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

`y'=-3x^2+12x-5\\y'=-6x+12\\\\`

Then substitute y' = 0 to get a vertex.

`y'=-6x+12\\0=-6x+12\\-12=-6x\\2=x`

Then substitute x = 2 in the equation of y = -3x²+12x-5

`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`

x = 2 and y = 7, therefore. The vertex is (2,7)