I need to find the vertex of each function.
y=-3(x-4)2 +6
do you mean
y=-3(x-4)^2 +6 ?????
if so, that function is symmetric about the vertical line x = 4
so
find y when x = 4 and you have it
recall the form of a quadratic equation:
y = a(x-h)^2 + k
where, (h,k) is the vertex,,
since the given equation is in that form (where a=3, h=4, and k=6), the vertex is at (4,6)
*there is an alternative solution, but this is used in higher math,, if you already know how to differentiate (or to get the derivative) of a function, you can use it to get the maximum or minimum,, in this case,
y=-3(x-4)^2 +6 *differentiate; set y=0
0=-6(x-4) *solve for x
x=4
then substitute it back to original equation, and you'll get 6,,
so there,, =)
To find the vertex of a function in the form of y = a(x - h)^2 + k, you need to identify the values of h and k. In this case, h represents the x-coordinate of the vertex and k represents the y-coordinate.
In your given function, y = -3(x - 4)^2 + 6, h = 4 and k = 6. Therefore, the vertex of the function is (4, 6).