What is the point slope of an equation with slope of 3/5 that passes through the point (10,-2)
2 answers:
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Answer:
The equation for rational function for given asymptotes is
f(x)=(-4x^2-6)/{(x-3)(x+3)}
Step-by-step explanation:
Given:
vertical Asymptotes at x=3 and x=-3 and a horizontal asymptote at
y=-4 i.e parallel to x axis.
To find:
equation of a rational function i.e function in form p/q
Solution;
the equation should be in form of p/q
Numerator :denominator.
Consider f(x)=g(x)/h(x)
as vertical asymptote are x=-3 and x=3
denominator becomes, (x-3) and (x+3)
for horizontal asymptote to exist there should have same degrees in numerator and denominator which of '2'
when g(x) will be degree '2' with -4 as coefficient and dont have any real.
zero.
By horizontal asymptote will be (-4x^2 -6)
The rational function is given by
f(x)=g(x)/h(x)
={(-4x^2-6)/(x-3)(x+3)}.
