So, I was doing some calculus application exercises and I can't get the following question:
To attack enemy positions, a fighter jet flies at low heights, going through the trajectory determined by the graph of the following equation:
$$f(x) = 1 + \frac{1}{x}, x > 0$$
a - Determine, by the derivative definition, the equation for the line tangent to $f(x)$ at a generic point $(a,f(a))$.
b - If a shot is made at the position $(1,2)$, determine at which point it hits the $x$-axis.
c - Determine the point on the graph of $f(x)$ at which a shot has to be made to hit a target at the point $(8,0)$.
I already got the derivative: $$f'(a) = -\frac{1}{a^2}$$ I know that I have to use the definition $$y - y_0 = m(x - x_0)$$ However, I cannot get past that.
Notice that if you use: $$y-y_0=m(x-x_0)$$ You should obtain: $$y-f(a)=f'(a)\cdot (x-a)$$ $$y-\left(1+\frac{1}{a}\right)=-\frac{1}{a^2}(x-a) \tag{1}$$ Equation $(1)$ is the general equation for the line tangent to $f(x)$ for a generic point $(a,f(a))$.
For (b), all you need to do is substitute $a=1$ into $(1)$ since the shot is made at $x=1$. Thus, you should obtain the tangent line at the point $(1,2)$.
For (c), you must find the value of $a$ such that the tangent line (Given by $(1)$) intersects with $(8,0)$. To do so, just substitute $x=8$ and $y=0$ into $(1)$, and find the positive value of $a$ for which this equality satisfies. That value of $a$ will be the $x$-coordinate of that point. Use $f(x)$ to find the $y$-coordinate of the point.