How do you find intercepts, extrema, points of inflections, asymptotes and graph #y=xsqrt(16x^2)#?
2 Answers
Explanation:
Consider:
The function is defined for
At the function is continuous in a compact domain, it cannot have asymptotes.
(1) We can find the intercepts solving the equation:
which yields three roots:
(2) Calculate the first derivative:
so
Solving the inequality:
we can see that:
This means that

decreasing in the interval
#x in [4,2sqrt(2)]# 
increasing in the interval
#x in [2sqrt(2),2sqrt(2)]# 
decreasing in the interval
#x in [2sqrt(2),4]#
and thus:
#x_m = 2sqrt(2)# is a local minimum#x_M = 2sqrt(2)# is a local maximum
Based on Weierstrass' Theorem, as
(3) We calculate the second derivative:
and we can see that
#f''(x) > 0# for#x in (4,0)# and so#f(x)# is concave up.#f''(x) < 0# for#x in (0,4)# and so#f(x)# is concave down.
graph{x sqrt(16x^2) [20, 20, 10, 10]}
See explanation.
Explanation:
To make y real,
xintercept ( y = 0 ) :
The graph also passes through the origin
Upon substitution
The graph is contained in an 8
question of search for asymptote(s) does not arise.
=0, at
Max y = 8, at
Min y =
=0, only when
the numerator ).
The origin is the point of inflexion ( after studying tangentcrossing 
curve, at
graph{(yxsqrt(16x^2))(y8.1)(y+8.1)=0[5, 5, 9, 9]}