Easy
y=a(x-h)^2+k
vertex is (h,k)
we know that vertex is (4,0)
input that point for (h,k)
y=a(x-4)^2+0
y=a(x-4)^2
passes thorugh the point (6,1)
input that point to find a
1=a(6-4)^2
1=a(2)^2
1=a(4)
divide both sides by 4
1/4=a
thefor the equation is
y=(1/4)(x-4)^2
or
y=(1/4)x^2-2x+4
Answer:
Algebraically, f is even if and only if f(-x) = f(x) for all x in the domain of f. A function f is odd if the graph of f is symmetric with respect to the origin. Algebraically, f is odd if and only if f(-x) = -f(x) for all x in the domain of f.
Step-by-step explanation:
Answer:
-7/11
Step-by-step explanation:
