For what value of the parameter #k# does the curve function #y=x^3+kx+k^2−6k+5# passes through the origin?

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Answer 1 · 2017-04-09T06:09:32

#k=1# and #k = 5#

Given: #y=x^3+kx+k^2−6k+5#

Substitute 0 for both x and y:

#0=0^3+k(0)+k^2−6k+5#

#0 = k^2−6k+5#

Factor

#(k-1)(k-5)= 0#

#k=1# and #k = 5#