The line is parallel to y= 2x + 1 and passes through point (-2,-1).
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Answer:
Step-by-step explanation:
f(x) = x2 + 2x - 2 should be rewritten using " ^ " to indicate exponentiation:
f(x) = x^2 + 2x - 2.
We find a couple of key points and use the fact that this parabola is symmetric about the line
-2
x = ----------- = -1. When x = -1, y = f(-1) = (-1)^2 + 2(-1) - 2, or 1 - 2 -2, or -3.
2(1)
Thus the vertex is at (-1, -3). The y-intercept is found by letting x = 0: y = -2. The axis of symmetry is x = -1.
Graph x = -1 and then reflect this y-intercept (0, -2) about the line x = -1, obtaining (-2, -2). If necessary, find 1 or two more points (such as the x-intercepts).
To find the roots (x-intercepts), set f(x) = x^2 + 2x - 2 = 0 and solve for x.
Completing the square, we obtain x^2 + 2x + 1 - 2 = + 1, or (x + 1)^2 = 3.
Taking the square root of both sides yields x + 1 = ±√3. One of the two roots is x = 1.732 - 1, or 0.732, so one of the two x-intercepts is (0.732, 0).
