The <em>quadratic</em> function g(x) = (x - 5)² + 1 passes through the points (2, 10) and (8, 10) and has a vertex at (5, 1).
<h3>How to analyze quadratic equations</h3>
In this question we have a graph of a <em>quadratic</em> equation translated to another place of a <em>Cartesian</em> plane, whose form coincides with the <em>vertex</em> form of the equation of the parabola, whose form is:
g(x) = C · (x - h)² - k (1)
Where:
- (h, k) - Vertex coordinates
- C - Vertex constant
By direct comparison we notice that (h, k) = (5, 1) and C = 1. Now we proceed to check if the points (x, y) = (2, 10) and (x, y) = (8, 10) belong to the parabola.
x = 2
g(2) = (2 - 5)² + 1
g(2) = 10
x = 8
g(8) = (8 - 5)² + 1
g(8) = 10
The <em>quadratic</em> function g(x) = (x - 5)² + 1 passes through the points (2, 10) and (8, 10) and has a vertex at (5, 1).
To learn more on parabolae: brainly.com/question/21685473
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Answer: i think no cuz f(1)=2 but i am not sure
Step-by-step explanation:
<span>tan(15) =
sin(15) / cos(15) =
sin(45 - 30) / cos(45 - 30) =
[ sin(45)cos(30) - sin(30)cos(45) ] / [ cos(45)cos(30) + sin(45)sin(30)]
Since sin(45) = cos(45) = √2/2, you can just factor that out from the top and bottom
[ cos(30) - sin(30) ] / [ cos(30) + sin(30)]
[ √3/2 - 1/2 ] / [ √3/2 + 1/2]
(√3 - 1) / (√3 + 1)
(√3 - 1)^2 / (√3+1)(√3 - 1)
(√3 - 1)^2 / (3 - 1)
(3 - 2√3 +1) / 2
2 - √3
There's also a formula for tan(a-b), but I couldn't remember it off hand.</span>
This is all I got. Hope this helps:
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