This equation has no solution
Answer: see below
<u>Step-by-step explanation:</u>
The vertex form of a quadratic equation is: y = a(x - h)² + k where
- "a" is the vertical stretch (positive = min [U], negative = max [∩])
- (h, k) is the vertex
- Axis of Symmetry is always: x = h
- Domain is always: x = All Real Numbers
- Range is y ≥ k when "a" is positive or y ≤ k when "a" is negative
a) y = 2(x - 2)² + 5
↓ ↓ ↓
a= + h= 2 k= 5
Vertex: (h, k) = (2, 5)
Axis of Symmetry: x = h → x = 2
Max/Min: "a" is positive → minimum
Domain: x = All Real Numbers
Range: y ≥ k → y ≥ 5
b) y = -(x - 1)² + 2
↓ ↓ ↓
a= - h= 1 k= 2
Vertex: (h, k) = (1, 2)
Axis of Symmetry: x = h → x = 1
Max/Min: "a" is negative → maximum
Domain: x = All Real Numbers
Range: y ≤ k → y ≤ 2
c) y = -(x + 4)² + 0
↓ ↓ ↓
a= - h= -4 k= 0
Vertex: (h, k) = (-4, 0)
Axis of Symmetry: x = h → x = -4
Max/Min: "a" is negative → maximum
Domain: x = All Real Numbers
Range: y ≤ k → y ≤ 0
d) y = 1/3(x + 2)² - 1
↓ ↓ ↓
a= + h= -2 k= -1
Vertex: (h, k) = (-2, -1)
Axis of Symmetry: x = h → x = -2
Max/Min: "a" is positive → minimum
Domain: x = All Real Numbers
Range: y ≥ k → y ≥ -2
What are the answer choices?
Answer:
A
Step-by-step explanation:
the end of interval is t=2
v=s'= -3t^2+4t-2
v(2)= -3(2^2)+4(2) -2
v(2)=-12+8-2
v(2)= -6m/s
so tje ans is A
<span>In this equation, 36>11 and so the ellipse has major axis along x axis
standard equation of ellipse with major axis is along x axis
x^2/a^2+y^2/b^2=1,
the foci are at (ae,0) and (–ae,0)
eccentricity e is given by
ae=√(a^2-b^2)
In this case ae=√(36-11)= √25=+/- 5
Focii are(-5,0) and(5,0) </span>
