prob:
Given that f(3)=-1 and f derivative(3)=5, find an equation for the tangent line to the graph of y=f(x) at x=3
the answer is y=5x-16 , but i dont know how to get there :/
any help is much appreciated!
prob:
Given that f(3)=-1 and f derivative(3)=5, find an equation for the tangent line to the graph of y=f(x) at x=3
the answer is y=5x-16 , but i dont know how to get there :/
any help is much appreciated!
Well the derivative is defined as the slope of the tangent line, so if you use slope-intercept notation of y=mx+b, then
you already know : y=5x + b for the tangent line.
Because the tangent line is at x=3, then a point on the tangent line is also at x=3. The problem gives you f(3)= -1 so there is a tangent line at (3,-1).
Plug those values into the equation to get b: -1=5(3) + b.
So, -1= 15 + b ........... -16=b!!
So the final equation is y=5x -16.
Glad I could use my calc knowledge somewhere! you can thank me later.....