1. Identify the type of conic section that has the equation 16x2 + 4y2 = 16 and identify its domain and range.
Canned answer: Wrong again!!!!
16x2 + 4y2 = 16
divide both sides by 16
x^2 + y^2 / 4 = 1
ellipse with centre (0,0) , the major axis along the y-axis
vertices at (0, ± 2) and (± 1, 0)
even if it had been a circle, the domain could not have been the real numbers ....
correct answer:
domain: all real numbers, such that -1 ≤ x ≤ 1
range : all real numbers such that -2, ≤ y ≤ 2
The Geometry to Algebra 2 Unit Test! (Questions vary, please read before answering)
1.
Identify the type of conic section that has the equation 16x2 + 4y2 = 16 and identify its domain and range. (1 point)
(Correct answer)
ellipse
Domain: {–1 ≤ x ≤ 1}
Range: {–2 ≤ y ≤ 2}
2.
Use the graph to write an equation for the parabola.
Image: graph - line curved as a u touching (0,0) and passing through points (3,3)(-3,3)
(Correct answer) y = (x^2/3)
3.
Find an equation of a parabola with a vertex at the origin and directrix y = –2.5
(Correct answer) y = 1/10 x^2
4.
What is the equation of the parabola with vertex (2, 4) and focus (5, 4)?
(Correct answer) x = 1/12 (y – 4)^2 + 2
5.
Write an equation of a circle with the given center and radius.center (3, 4) and radius 6
(Correct answer) (x – 3)^2 + (y – 4)^2 = 36
6.
Write an equation for the translation of x^2 + y^2 = 25 by 8 units left and 4 units down.
(Correct answer) (x + 8)^2 + (y + 4)^2 = 25
7.
Write an equation in standard form for the given circle. Image: graph - no points but the arcs do almost pass on point (0,-1) and (0,-5)
(Correct answer) (x + 1)^2 + (y + 3)^2 = 4
8.
Identify the center and radius of a circle with equation (x - 4)^2 + (y + 2)^2 = 9.
(Correct answer) Center at (4,-2); radius: 3
9.
Write an equation of the ellipse with foci at (0, ±10), and vertices at (0, ±11).
(Correct answer) x^2/21 + y^2/121 = 1
10.
What are the foci of the ellipse given by the equation 25x^2 + 16y^2= 400?
(Correct answer) (0, ± 3)
11.
How far apart are the foci of an ellipse with a major axis of 34 feet and a minor axis of 16 feet?
(Correct answer) 30 feet
12.
What are the foci of the hyperbola with equation 5y^2 – 4x^2 = 20?
(Correct answer) (0, ± 3)
The final score is 12/12 (100%).
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To identify the type of conic section that has the equation 16x^2 + 4y^2 = 16, we can analyze the coefficients of the x^2 and y^2 terms.
Since the coefficients of x^2 and y^2 are positive and equal, it indicates an ellipse. The general equation for an ellipse centered at the origin is (x^2/a^2) + (y^2/b^2) = 1, where 'a' represents the semi-major axis and 'b' represents the semi-minor axis.
Comparing this equation to the given equation, we can observe that the semimajor axis (a) = sqrt(16) = 4 and the semiminor axis (b) = sqrt(4) = 2.
Hence, the equation 16x^2 + 4y^2 = 16 represents an ellipse with a horizontal major axis of length 8 (2a) and a vertical minor axis of length 4 (2b).
Now, let's determine the domain and range of the ellipse.
For the x-coordinate (domain), we want to find the values of x that satisfy the equation. In this case, since it's a complete ellipse, the x-coordinates extend from -4 to 4.
So, the domain of the ellipse is -4 ≤ x ≤ 4.
For the y-coordinate (range), we can rewrite the given equation as y^2 = 4 - (x^2/4) to isolate y. Taking the square root on both sides, we have y = ± sqrt(4 - (x^2/4)).
However, since it's an ellipse, the range depends on the value of x within the domain. As x varies from -4 to 4, the y-values will vary accordingly. For the given equation, the range of the ellipse extends from -2 to 2.
Therefore, the range of the ellipse is -2 ≤ y ≤ 2.