Carcass wrote:
Attachment:
GRe parabola.jpg
Which is an equation of the preceding graph?
Indicate all such equations.
(A) \(y = –x^2 + 4x + 4\)
(B) \(y = –(x + 2)^2\)
(C) \(y = (x – 2)^2\)
(D) \(y = x^2 – 4x + 4\)
(E) \(y = (x + 2)^2\)
(F) \(y = x^2 + 4x + 4\)
For this kind of question, a quick approach is to apply the following extremely important property..
If a point lies ON a graph, then the x- and y-coordinates of that point must satisfy the equation of that graph. So, for example, we can see that the point (-1, 1) lies ON the given graph.
This means \(x = -1\) and \(y = 1\) must be a solution to the equation of the graph.
Let's plug \(x = -1\) and \(y = 1\) into each answer choice to see whether it's a solution to that equation. We get:
(A) \(1 = –(-1)^2 + 4(-1) + 4\)
Evaluate to get:
\(1 = -1\). Since \(x = -1\) and \(y = 1\) is NOT a solution to this equation,
we can eliminate this answer choice. (B) \(1 = –((-1)+ 2)^2\)
Evaluate to get:
\(1 = -1\). Since \(x = -1\) and \(y = 1\) is NOT a solution to this equation,
we can eliminate this answer choice. (C) \(1 = ((-1) – 2)^2\)
Evaluate to get:
\(1 = 9\). Since \(x = -1\) and \(y = 1\) is NOT a solution to this equation,
we can eliminate this answer choice. (D) \(1 = (-1)^2 – 4(-1) + 4\)
Evaluate to get:
\(1 = 9\). Since \(x = -1\) and \(y = 1\) is NOT a solution to this equation,
we can eliminate this answer choice. (E) \(1 = ((-1) + 2)^2\)
Evaluate to get:
\(1 = 1\). Since \(x = -1\) and \(y = 1\) IS a solution to this equation.
(F) \(1 = (-1)^2 + 4(-1) + 4\)
Evaluate to get:
\(1 = 1\). Since \(x = -1\) and \(y = 1\) IS a solution to this equation.
Since \(y = (x + 2)^2\) and \(y = x^2 + 4x + 4\) are equivalent equations, the correct answer here is E and F.