The zeros of a quadratic function are 2 and 4 what is the vertex of the function

The x-coordinate of the vertex is precisely halfway between the given roots 2 and 4; thus, x = (2+4)/2, or x=3.  The y-coordinate is the value of the function at this x=3.  

We can actually determine the equation of this quadratic from the zeros:

f(x) = (x-2)(x-4), or   f(x) = x^2 - 6x + 8.  To find the y-coord. of the vertex,, subst. 3 for x in f(x) = x^2 - 6x + 8:    f(3) = 3^2 - 6(3) + 8 = 9 - 18 + 8, or -1.  Then we know that the vertex is at (3, -1).

plz yuo

step-by-step explanation:

b) k’(6, 4); l’(3, 3)

step-by-step explanation:

a 270° clockwise rotation maps every point (x, y) to (y, -x). (it negates the x-coordinate and then switches it and the y-coordinate.)

this means the endpoints of the side congruent to kl, k' and l', would be mapped as:

k'(-4, 6)→(6, 4)

l'(-3, 3)→(3, 3)