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
Find the equation of a straight line that passes through a point (1,5) and tangent to curve y = x^3
3 answers
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.
y = x^3*
1 answer