Hi. I was unsure of what exactly you wanted from this equation, so here's a quick analysis:
<em>f(x) = 2(x - 3)^2 - 2</em>
<em></em>
Domain: (-∞, ∞)
Range: (-2, ∞)
X-intercepts: (4, 0), (2, 0)
Y-intercept: (0, 16)
Axis of Symmetry: x = 3
Minimum value (vertex): (3, -2)
Standard form: y = 2x^2 - 12x + 16
Check the picture below.
let's recall that once we move an angle in the opposite direction of the 0-line or x-axis, we end up with a negative counterpart angle.
since we know that θ is on the IV Quadrant, then its negative counterpart must be across the line, and tangent is negative on the IV Quadrant, and positive on the I Quadrant.
11, add six and a half to four and a half.