I was just introduced to derivatives.
I have a problem,
Find the slope of the tangent line to the graph at the indicated point.
The question was,
$$y=x^2 -4x -5 ; (-2,7)$$
So the indicated points is (-2,7)
Now, here's what I did (I tried finding the equation for $(x_1,y_1)$ first:
$$\lim_{\Delta x\to0} {f(x_1+\Delta x) – f(x_1)\over \Delta x}$$
$$$$
$$f(x)=x^2-4x-5$$
$$f(x_1)=x_1^2-4x_1-5$$
$$f(x_1+\Delta x) = (x_1+\Delta x)^2 -4(x_1+\Delta x) -5$$
$$$$
$$Now,$$
$$\lim_{\Delta x\to0} {(x_1+\Delta x)^2 -4(x_1+\Delta x) -5 – (x_1^2-4x_1-5)\over \Delta x}$$
$$\lim_{\Delta x\to0} {x_1^2+2x_1\Delta x+\Delta x^2 -4x_1 -4\Delta x -5 -x_1^2 +4x_1 +5\over\Delta x}$$
$$\lim_{\Delta x\to 0} {2x_1\Delta x +\Delta x^2 -4\Delta x\over \Delta x}$$
$$\lim_{\Delta x \to 0} 2x_1 +\Delta x -4$$
$$=2x_1 -4$$
I came up with the equation of the tangent line at the point $(x_1,y_1)$
Now, how do I turn it into equation of tangent line at point (-2,7)?
Or, was there a way to find the equation stragith to (-2,7) instead of having to find equation for $(x_1,y_1)$??
Best Answer
Your function is $f(x)=x^2-4x-5$, and you found its derivative: $f'(x) = 2x-4$ .
This is NOT the equation of the tangent line at any point $x$. The derivative just returns the value of the slope of the tangent line at any point $x$.
This means that if you want the slope of the tangent line of $f(x)$ at $(-2,7)$ (which is what you're asked) you just plug in $x=-2$ in the equation of the derivative: $$f'(-2)=2\cdot (-2)-4=-8$$
Notice that $f(-2)=7$, otherwise the point would not belong to the graph of the function $f(x)$.
Now you have the slope ($-8$) and a point $(-2,7)$, which means you can find the equation of the line at $x=-2$ with the formula: $$y=f(-2)+f'(-2)\cdot (x-(-2))$$ $$y=7-8(x+2)$$
This is the equation of a line given its slope $m$ and a point $(x_0,y_0)$: $y=y_0+m(x-x_0)$.
Now you were also wondering if there was any shortcut to the result. You've just been introduced to derivatives, and you still calculate them as limits, but you will see later that there are easier ways to calculate a derivative (not involving limits) which makes things extremely more simple (in this case it was a polinomial, but try to do it witha more complex function and you'll see!).
Once you learn tricks to differentiate, this kind of problems will take you a cuple of seconds.
On the other hand, you could have taken $x_1=-2$ (in your work) and it would have worked just as fine. But, in general, we calculate the derivative at any point $x$ and then evaluate it at the point of interest.
Hope this clarifies your thoughts a little bit.