Respuesta :

apologiabiology
for f(x)=ax²+bx+c
if a is positive, then it is concave up and the verx is minimum
if a is ngative then it is concave down and vertex is maximum

hack: in form f(x)=ax²+bx+c, the x value of vertex is -b/2a
to find y value, just subsitutte that value for x in f(x)

so
y=-1x²+4x-4
negative, so vertx is max
x value of vertex is -4/(2*-1)=-4/-2=2
find f(2) or subsitute 2 for x
y=-1(2)²+4(2)-4
y=-1(4)+8-4
y=-4+4
y=0

maximum is 0

A is answer
Y= -x^2+4x-4
Standard form is
Y=ax^2+bx+c

here cofficient of a is negative so this is downward parabola.
so it will have maximum value at its vertex
h=-b/2a
h= -4/2(-1)
h=2
put h in equation y= -x^2+4x-4
that will be other vertex K
k=-(2)^2+4*2-4
k=-4+8-4
k=0
vertex(h,k)=(2,0)
y=(x-2)^2+0
maximum value is k which is 0

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