How do you find intercepts, extrema, points of inflections, asymptotes and graph #h(x)=xsqrt(9-x^2)#?

Question

Christopher Allers · Accepted Answer

Please see below.

Domain : #[-3,3]# Intercepts #y# intercept #h(0) = 0# #x# intercepts: #-3#, #0#, and #3# (Solve #h(x) = 0#) Asymptotes The function has no horizontal asymptotes. (The domain is bouinded.) and no vertical asymptotes (there is no division). #h'(x) = (9-2x^2)/sqrt(9-x^2)# Is undefined only at the endpoints of the domain and is #0# at #x = +-3/sqrt2# It is worth noting that as #x# approaches the endpoints of the domain, #+-3#, the derivative goes to #+-oo# So the tangent line becomes vertical. Local extrema and increasing/decreasing On #[-3,-3/sqrt2)#, we have #h'(x) < 0# so #h# is decreasing. On #(-3/sqrt2,0)#, we have #h'(x) > 0# so #h# is increasing. And #h(-3/sqrt2) = -9/2# is a local minimum. On #(0,3/sqrt2)#, we have #h'(x) > 0# so #h# is increasing. On #(3/sqrt2,3]#, we have #h'(x) ><0# so #h# is decreasing. And #h(3/sqrt2) = 9/2# is a local maximum. Concavity and inflection points #h''(x) = (x(2x^2-27))/(9-x^2)^(3/2)# Is undefined only at the endpoints of the domain and is #0# at #x = 0# (The expression is also #0# at #x = +-sqrt(27/2)#, but those are outside the domain of #h#.) On #[-3,0)#, we have #h''(x) > 0# so the graph of #h# is concave upwards (convex). On #[-3,0)#, we have #h''(x) < 0# so the graph of #h# is concave downwards (concave). The point #(0,0)# is the only inflection point. The graph is: graph{xsqrt(9-x^2) [-9.23, 13.27, -5.625, 5.625]}

Luke Alvarez · Answer

undefined To find intercepts, set $ h(x) = 0 $ and solve for $ x $. To find extrema, take the derivative of $ h(x) $, set it equal to zero, and solve for $ x $. To find points of inflection, take the second derivative of $ h(x) $, set it equal to zero, and solve for $ x $. To find asymptotes, analyze the behavior of $ h(x) $ as $ x $ approaches positive or negative infinity.