ASAP!! Please help me. I will not accept nonsense answers, but will mark as BRAINLIEST if you answer is correctly with solutions.

second one

Step-by-step explanation:

The function g has this expression g(x) = ax²+c

To find when g reaches it's minimum we must derivate it

solve g'(x) = 0

replace x with 0 in g(x)

g(0) = a(0)²+c

g(0) = c

the maximum of f(x) is 1 so for g(x) to have a grather minimum c should be greather than 1

the second statement is true

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this

the square root of what number? if you add that, then someone will probably answer the question.