We are given
Vertical asymptotes:
Firstly, we will factor numerator and denominator
we get
We can see that (x-3) is common in both numerator and denominator
so, we will only set x+3 to 0
and then we can find vertical asymptote
Hole:
We can see that (x-3) is common in both numerator and denominator
so, hole will be at x-3=0
Horizontal asymptote:
We can see that degree of numerator is 2
degree of denominator is also 2
for finding horizontal asymptote, we find ratio of leading coefficients of numerator and denominator
and we get
y=1
now, we can draw graph
Graph:
The hyperbolic cos (cosh) is given by
cosh (x) = (e^x + e^-x) / 2
The slope of a tangent line to a function at a point is given by the derivative of that function at that point.
d/dx [cosh(x)] = d/dx[(e^x + e^-x) / 2] = (e^x - e^-x) / 2 = sinh(x)
Given that the slope is 2, thus
sinh(x) = 2
x = sinh^-1 (2) = 1.444
Therefore, the curve of y = cosh(x) has a slope of 2 at point x = 1.44
p(x)= x-2
g(x)= 2x^3 + 3x^2 - 11x - 6
first we have to find the zero of the polynomial of x-2
p(x)= x-2 = 0
x=2
therefore,
p(x)= 2x^3 + 3x^2 - 11x - 6
p(2)= 2*2^3 + 3*2^2 - 11*2 - 6
= 2*8 + 3*4 - 11*2 - 6
= 16 + 12 - 22 - 6
= 28-28
= 0
Hope it helped u, ^_^.
