Answer:
D. Minimum at (3, 7)
Step-by-step explanation:
We can add and subtract the square of half the x-coefficient:
y = x^2 -6x +(-6/2)^2 +16 -(-6/2)^2
y = (x -3)^2 +7 . . . . . simplify to vertex form
Comparing this to the vertex for for vertex (h, k) ...
y = (x -h)^2 +k
We find the vertex to be ...
(3, 7) . . . . vertex
The coefficient of x^2 is positive (+1), so the parabola opens upward and the vertex is a minimum.
Answer:
Option A. (-1, 0)
Step-by-step explanation:
In the figure attached,
Circle O is a unit circle (having radius r = 1 unit)
If a point A with central angles = θ, is lying on the circle then the coordinates of the point A will be,
x = r.cosθ
x = 1.cosθ = cosθ
and y = r.sinθ
y = 1.sinθ = sinθ
Therefore, coordinates representing the point A will be (cosθ, sinθ).
As per question the given point A is lying at P (a point having central angle θ = 180°)
Coordinates of point P will be
(x', y') → (cos180°, sin180°)
→ (-1, 0)
Therefore, Option A will be the answer.
