Question

find k such that y=ksqrt(x) is tangent to y=x+6

Answer 1

calculus is the best process to solve this question

Answer 2

The procedure is very much like your last question.

Answer 3

the derivative of the second equation is one. is the derivative of the first 1/2x?

Answer 4

CALCULUS WAY\[m=\frac{ k }{ 2\sqrt{x} }\]
\[\frac{ k }{ 2 \sqrt{x} }(1)=-1\]
\[k=-2\sqrt{x}\]

Answer 5

\[x+6=(-2\sqrt{x})\sqrt{x}\]
\[x+6=-2x\]
\[x=2,k \rightarrow -2 \sqrt{2}\]

Answer 6

that's not the answer but thank u

Answer 7

i think the question is supposed to be
"find k such that y=x+6 is tangent to y=ksqrt(x) "