The slope of the line tangent to [tex]y=x^3+2[/tex] at [tex]x=c[/tex] is [tex]3c^2[/tex]. If [tex]c=3[/tex], then the slope is [tex]3\cdot3^2=27[/tex].
Use the point-slope formula to get the equation of the tangent line passing through (3, 29), which we know has slope 27:
[tex]y-29=27(x-3)\implies y=27x-52[/tex]
