Problem: Find an equation tangent to the graph of y=f(x) at the point
where x=-3 if f(-3)=2 and f'(-3)=5
What I've tried:
I tried solving this the way "normal" tangent equations are found when the problem gives a point and an equation (y=Mx+b).[Using slope-intercept to then derivate]
I attempted finding Y1 and X1 (as one would do to solve for a point using Y-Y1=m(x-x1))
Where x1 =-3 [As given by the problem]
[The only equation given is] Y=f(x) [plugin-in "-3" results in] f(-3)) = Y
The problem establishes that f(-3)=2
Therefore, we have found y=2 & x1=-3
If I were to attempt to substitute this in order to "derivate" (as in a normal problem where asked to find a tangent and a point is given and solved trough slope intercept, then trough derivation):
y=f(x)
y'=f'(x)
f'(-3) = 2 //How could I derivate this?, here is probably my mistake, though I do not know how to approach it in another way.
[Then the problems establishes "and f'(-3)=5" so according to the problem f'(-3)=5 NOT 2, therefore I did something wrong and clearly can not be solved this way]
I am completely stuck here really. How can I find the equation?
Best Answer
The tangent has the slope $f'(-3)=5$ and passes through the point $(-3,f(-3))=(-3,2)$. Take $y=Mx+b$ and substitute $M=f'(-3)=5$ and the tangent becomes $y=5x+b$. Now, substituting $(-3,2)$ gives $b=2-5(-3)=17$. So, the tangent is $y=5x+17$.
The slope of this tangent would be $\frac{dy}{dx}=5$ which is the same as the slope of the curve $f(x)$ at the point $(-3,2)$.