(9x^2+8x)-(2x^2+3x)
9x^2+8x-2x^2-3x
7x^2+5x
Given:
The function is

To find:
The asymptotes and zero of the function.
Solution:
We have,

For zeroes, f(x)=0.



Therefore, zero of the function is 0.
For vertical asymptote equate the denominator of the function equal to 0.


Taking square root on both sides, we get


So, vertical asymptotes are x=-4 and x=4.
Since degree of denominator is greater than degree of numerator, therefore, the horizontal asymptote is y=0.
Answer:
(5, 11) and (2, 2)
Step-by-step explanation:
y = (x-2)² + 2
y + 4 = 3x
(x-2)² + 2 + 4 = 3x
x² - 4x + 4 + 6 = 3x
x² - 7x + 10 = 0
(x - 5)(x - 2) = 0
x - 5 = 0, x = 5
x - 2 = 0, x = 2
y = (5-2)² + 2 = 11
(5, 11)
y = (2-2)² + 2 = 2
(2, 2)
The answer is Vertical angles are equal.