Use technology to construct an appropriate model of the data. (0,1), (1,-2), (2,-3), (3,-2), (4,1) f(x) = – x2 + 4 f(x) = x2 - 4
f(x) = - x2 + 4x - 1 f(x) = x2 - 4x + 1
a. f(x) = – (x+2)2 + 3B. f(x) = (x+2)2 + 3C. f(x) = – (x-2)2 - 3D. f(x) = (x-2)2 - 3
a. f(x) = – (x+2)2 + 3B. f(x) = (x+2)2 + 3C. f(x) = – (x-2)2 - 3D. f(x) = (x-2)2 - 3
1 answer:
The answer choices tell you this is a parabola.
The given points tell you it opens upward and the vertex is (2, -3).
Your knowledge of the vertex form of the equation is the only "technology" that you need. The vertex form is
y = a(x - h)² + k . . . . . for vertex (h, k)
The fact that the y-value is 1 more than its vertex value when |x-2| = 1 tells you the scale factor "a" is 1.
You can write the model as
D. f(x) = (x -2)² - 3
The given points tell you it opens upward and the vertex is (2, -3).
Your knowledge of the vertex form of the equation is the only "technology" that you need. The vertex form is
y = a(x - h)² + k . . . . . for vertex (h, k)
The fact that the y-value is 1 more than its vertex value when |x-2| = 1 tells you the scale factor "a" is 1.
You can write the model as
D. f(x) = (x -2)² - 3
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Answer: 24.2° SouthWest
<u>Step-by-step explanation:</u>
First step: DRAW A PICTURE of the vectors from head to tail <em>(see image)</em>
I created a perpendicular from the resultant vector to the vertex of the given vectors so I could use Pythagorean Theorem to find the length of the perpendicular. Then I used that value to find the angle of the plane.
<u>Perpendicular (x):</u>
cos 35° = adjacent/hypotenuse
cos 35° = x/160
→ x = 160 cos 35°
<u>Angle (θ):</u>
sin θ = opposite/hypotenuse
sin θ = x/320
sin θ = 160 cos 35°/320
θ = arcsin (160 cos 35°/320)
θ = 24.2°
Direction is down (south) and left (west)
