finding a curve tangent to a line I am studying Strang's Calculus text and find myself stumped by the following problem:
'Find a curve that is tangent to y = 2x - 3 at x = 5. Find the normal line to that curve at (5, 7).'
All my searching only yields results for finding tangent lines to a curve, not the other way around.
I thought to take the antiderivative of the function and substitute the point of tangency back into the equation to solve for the resulting constant, but this didn't give me the correct answer.
Suggestions on how to proceed?
 A: They want "a curve" - because there are an infinite number of curves that will satisfy the conditions!
So you need $y=7$ when $x=5$ and $\frac {dy}{dx}=2$ when $x=5$.
The simplest kind of curve would be a quadratic; I would suggest $y=x^2+ax+b$.
Substitute the known conditions into that curve to create simultaneous equations for $a$ and $b$.
Solve and there you are!
A: Choose any function $f(x)$ such that sensible values for $f(5)$ and $f'(5)$ exist.
We want to create a new function $g(x)$ such that $g(5)=7$ and $g'(5)=2$.
Let $g(x)=af(x)+b \Rightarrow 7=af(5)+b$
Differentiation gives $g'(x)=af'(x) \Rightarrow 2=af'(5)$
So $a=\frac 2 {f'(5)}$
And $7=\frac 2 {f'(5)}f(5)+b$
So $b=7-\frac {2f(5)} {f'(5)}$ 
Which gives $g(x)=\frac {2f(x)} {f'(5)}+7-\frac {2f(5)} {f'(5)}$
Or $g(x)=\frac {2f(x)} {f'(5)}-\frac {2f(5)} {f'(5)}+7$
A: HINT
Take any curve involving two constants.
Find two intersection points with the given line
$$ \frac{x}{3/2}+\frac{y}{3}+=1 $$
by solving the two together.
Relate the two constants so the points coincide  to a single tangent point, involving only a single constant.
