What is the vertex of the graph of y = 2(x – 3)^2 + 4?
A.) (4, 3)
B.) (4, –3)
C.) (3, 4)
D.) (–3, 4)
Solve. (x + 2)2 = 1
A.) x = –3 and x = –1
B.) x = 3 and x = 1
C.) x = –1 and x = 4
D.) x = –4 and x = –3
Please help. I cant figuire out these problems. Best answer !
What is the range of the graph of y = 5(x – 2)^2 + 7?
A.) y ≥ 7
B.) y ≤ 7
C.) y ≥ 2
D.) y ≤ 2
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(1) The equation is in completed square form and we know immediately that the vertex is at x = 3 and hence y = 4.
Ans: C (3,4)
(2) Assume this is (x+2)^(2) = 1
=> (x+2) = +/- sqrt(1) = +/- 1
=> x = +/-1 - 2
=> x -3 or x = -1
Ans (A)
(3) The range of y = 5(x-2)^2 + 7 is the set of values that y takes. We start by noting that the domain is the set of values that x can take and there are no restriction. That is x can be any real number. As per (1) we note that the vertex is at x = 2 and this is when y takes it's minimum value.. Hence:
y = 5(2-2) + 7
=> y = 7
Hence the range is y >/= 7
Ans: A
A.
A.
A. Since a is not negative, and y is 7.