3x² + 5x - 2 = 0
3x² + 6x - x - 2 = 0
3x(x) + 3x(2) - 1(x) - 1(2) = 0
3x(x + 2) - 1(x + 2) = 0
(3x - 1)(x + 2) = 0
3x - 1 = 0 or x + 2 = 0
+ 1 + 1 - 2 - 2
3x = 1 or x = -2
3 3 1 1
x = ¹/₃ or x = -2
f(x) = 3x² + 5x - 2
f(¹/₃) = 3(¹/₃)² + 5(¹/₃) - 2
f(¹/₃) = 3(¹/₉) + 1²/₃ - 2
f(¹/₃) = ¹/₃ - ¹/₃
f(¹/₃) = 0
(x, f(x)) = (¹/₃, 0)
f(x) = 3x² + 5x - 2
f(-2) = 3(-2)² + 5(-2) - 2
f(-2) = 3(4) - 10 - 2
f(-2) = 12 - 12
f(-2) = 0
(x, f(x)) = (-2, 0)
Vertical Asymptotes: ¹/₃ or -2
Horizontal Asymptotes: 0
Oblique Asymptote: No Asymptotes
Slope would be undefined because there are many many y-values for one x value. This is impossible for ANY function. To be specific, the correct equation that names the graph is <em>x=4</em>. Please note that this is <u>not</u> a function. If it goes straight up, it is <u>UNDEFINED</u>!!!
Hope I helped!
The answer is <u><em>D</em></u>
N + 2 = 13
-2 -2
n = 11
I hope this helps!
Answer:
Sally is not right
Step-by-step explanation:
Given the two sequences which have their respective 
terms as following:
Sequence A. 
Sequence B. 
As per Sally, there exists only one number which is in both the sequences.
To find:
Whether Sally is correct or not.
Solution:
For Sally to be correct, we need to put the terms of the respective sequences as equal and let us verify that.
When we talk about terms, 
here is a whole number not a fractional number.
But as per the statement as stated by Sally is a fractional number, only then the two sequences can have a number which is in the both sequences.
Therefore, no number can be in both the sequences A and B.
Hence, Sally is not right.
If its a reflection across the y axis the x value is reflected and the y value stays the same, so (-3,7)
