Asked by asdf

Find the equation of a straight line that passes through a point (1,5) and tangent to curve y = x^3
Answered by domulus
f(x)=x^3
f'(x)=3x^2
m=f'(1)=3(1)^2=3

y=mx+b where y=5, m=3, x=1
5=3*1+b
b=2
so equation of straight line is
y=3x+2
Answered by asdf
The point (1,5) isn't on the graph of y^x^3. It is just a point on the tangent of the graph y = x^3. Sorry if I wasn't clear before.
Answered by asdf
y = x^3*
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