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Currently reading Gelfand's Functions and Graphs as part of pre-calculus and stuck on this question. The author discusses a way to find a tangent to a curve without the use of limits or derivatives with the following sample solution:

  1. Find the tangent at the point $O(0, 0)$ to the parabola $y=x^2+x$.

Solution: Let us take some point $M$ on the parabola with coordinates $(a, b)$. Obviously, $b = a^2+a$. Let us draw a straight line through the points $O$ and $M$. The equation of this straight line has the form $y = kx$. At $x=a$, we have $y=a^2+ a$, hence $k = a+1$, and the equation of the secant is $y =(a+1)x$. We shall now make the point $M(a,b)$ approach the point $O(0, 0)$. When the point $M$ coincides with the point $0$, its abscissa $a$ vanishes, and the secant $y=(1+a)x$ becomes the tangent $y=x$.

Now he asks us to find the tangent at a different point of the same curve:

  1. Find the tangent to the parabola $y = x^2+x$ at the point $A(1, 2)$.

It is clear that the tangent at $A(1,2)$ will have the general form $y=mx+c$ since it does not pass through $O$ and will at $(1,2)$ be of the form $2=m+c$. How do we find the values of $m$ and $c$ here?

studyingsicp
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    For the first solution, in my opinion, when the author uses the words "approach" and "vanishes", he basically uses limit without explicitly saying so. – acat3 Jan 09 '24 at 15:01
  • Would it be possible to find the tangent without using the $dy/dx$ formula. I am trying to self study this as part of pre-calculus and just wanted to know a way to find the solution without involving calculus. All the available solutions solve it through differentiation. – studyingsicp Jan 09 '24 at 16:07
  • Yes, it's possible, by using parabola's locus definition. Look up focus and directrix of a parabola, then try to find a way to formulate an expression of the tangent line of a parabola. – acat3 Jan 09 '24 at 16:13

3 Answers3

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There's a very simple principle for that: when the parabola $y=P(x)$ has a tangent line of the form $y=mx+q$, then the equation $y=P(x)-(mx+q)$ has two roots which are equal.

So, in your case: as $y=mx+q$ has $(1,2)$ as a point, you know that$m+q=2$, giving you a relationship between $m$ and $q$.

Next: $P(x)-(mx+q)$ can be written as $x^2+x-(mx+2-m)$, or shortly: $x^2+(1-m)x+(m-2)$, who needs to have two real roots, meaning that the discriminant must be zero.

Dominique
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  • Understood. Since the tangent is essentially a double point of coincidence, solving for the discriminant gives the solution m=3 – studyingsicp Jan 11 '24 at 06:38
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Adding a solution here that seems to be right, in case this is useful for someone else: Let $M(a,a^2+a)$ be another point in the parabola and we consider $AM$ as the secant line.

At $M$, the secant line equation will be of form: $a^2+a = ma + c$

Since we also have $(1,2)$ in this secant, its equation at that point becomes $2 = m+c$.

Subtracting these two line equations, $(a^2+a = ma + c ) - ( 2 = m +c)$ gives $m = (a+2)$.

Now the secant's line equation becomes $y = (a+2)x+c$. Solving for $c$ from this equation by substituting the point $A(1,2)$ gives $2 = (a+2) + c$ or $c = -a$.

Hence the secant now becomes $y = (a+2)x - a$

As $M$ approaches $A$, the number $a$ approaches $1$, so the secant line closely approximates the line with equation $y = (1+2)x - 1$ or $y = 3x-1$, which is the tangent at $A(1,2)$.

studyingsicp
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If you draw a picture you'll see that any tangent of the parabola doesn't meet the parabola anywhere else (think about what kinds of functions have this property). In order for the parabola and the tangent to meet, this has to be satisfied: $$ mx+c=y=x^2+x $$ Manipulate this equation a little bit to bring it to the familiar form of a second order equation: $$ x^2 + (1-m)x-c = 0 $$ You want this equation to only have one (real) solution, since otherwise the line will meet the function plot at two places. You should know that this means the discriminant is equal to 0. Setting the discriminant to be 0 will give you a relation between $m$ and $c$ (let's say it gives you $m$ as a function of $c$), and then you plug that back into the condition for the two functions to meet and solve for $c$. That gives you the result!

Andreas Christophilopoulos
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