# C o n c e p t s

This topic requires familiarity with the following concepts:

• Integration by parts
• Differentiation
• Differential equations
• Partial fractions

The intercept formulas are formulas that give the point at which a tangent of a function f(x) crosses the x-axis or y-axis. These formulas are nothing new. I just gave them a naming scheme. I denoted y-intercept formula as fyi(x) or fyi or yyi. The x-intercept formula is denoted fxi(x) or fxi or yxi. So a tangent to f(x) at (p, f(p)) crosses the y-axis at (0, fyi(p)) and x-axis at (fxi(p), 0). One application of the x-intercept formula is Newton's Method for finding the real zeros of a polynomial.

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As I mentioned that these formulas are not new. I just formally gave them names and characterized them as functions. The slope of a tangent line at (p, f(p)) to a curve f(p) is f'(p). Using the point-slope equation of a line, we can find the y-intercept and x-intercept functions.

$y - {y}_{1} = m(x - {x}_{1})$ or $y-f(p)=f'(p)(x-p)$

Letting x equal zero, we obtain ${{y}_{yi}}=f(p)-p\cdot f'(p)$, the y-intercept function of f. The notation for the y-intercept for f(x) is fyi(x), in terms of x. Letting y equal 0, we find the x-intercept function of f(x) to be ${{f}_{xi}}\text{(}x)=-\frac{{{f}_{yi}}(x)}{f'(x)}$.

y-Intercept Formula. For a function f(x), the y-intercept formula is ${{f}_{yi}}(x)=f(x)-x\cdot f'(x)$, where fyi(x) gives the y value of the point where a tangent to the curve at (x, f(x)) crosses the y-axis or (0, fyi(p)).

x-Intercept Formula. For a function f(x), the x-intercept formula is ${{f}_{xi}}\text{(}x\text{) }=\text{ }-\frac{{{f}_{yi}}(x)}{f'(x)}\text{ }=\text{ }x-\frac{f(x)}{f'(x)}$, where fxi(x) gives the y value of the point where a tangent to the curve at (x, f(x)) crosses the x-axis or (fxi(p), 0).

## Newton’s Method

When we search for the zeros of a polynomial function, we are looking for the point where the function crosses the x-axis. We can use the x-intercept formula for approximating zeros of a polynomial equation to any degree of accuracy. We can also use it for other functions, such as transcendental functions; however, that would require more than a paper and pencil. The x-intercept formula can also be written as $\frac{x\cdot f\text{ }\!\!'\!\!\text{ (}x\text{)}-f\text{(}x)}{f\text{ }\!\!'\!\!\text{ (}x\text{)}}$. Using an initial value for x, we can approximate a zero r1 which may be close to x. This initial value for x will give the x-intercept of the tangent at x. But using the value obtained we can find the x-intercept again of the tangent and be closer to the zero. Repeating this procedure will bring the x-intercept closer and closer to the zero. Let’s use an example to illustrate the method using the function ${{x}^{2}}-4=0$. We already know the zeros to be 2 and –2.

Let ${{r}_{1}}\approx x-\frac{f\text{(}x\text{)}}{f\text{ }\!\!'\!\!\text{ (}x\text{)}}=x-\frac{{{x}^{2}}-4}{2x}=\frac{{{x}^{2}}+4}{2x}$. We know one of the roots is 2 so we’ll choose an initial x value close to 2. Let’s say we choose a 6. Then, ${{r}_{1}}\approx \frac{{{6}^{2}}+4}{2(6)}=\frac{10}{3}$. Now, we iterate the process:

${{r}_{2}}\approx \frac{{{({10}/{3}\;)}^{2}}+4}{2({10}/{3}\;)}=\frac{34}{15}$, ${{r}_{3}}\approx \frac{{{({34}/{15}\;)}^{2}}+4}{2({34}/{15}\;)}=\frac{514}{255}$, ${{r}_{4}}\approx \frac{{{({514}/{255}\;)}^{2}}+4}{2({514}/{255}\;)}=\frac{131074}{65535}\approx \text{2}\text{.00006}$

We can see a pattern emerge: ${{r}_{n}}=\frac{2\cdot {{2}^{{{2}^{n}}}}+4}{{{2}^{{{2}^{n}}}}}$. The limit of this expression is 2. Sometimes, the values jump around instead of converging. In this case, you have to choose some other starting value.

## Implicit Functions

The intercept formulas can also be used for implicit functions. Because we can find the slope through implicit differentiation, we can also find the x- and y-intercepts implicitly. Since the formula will not be a function, we’ll denote the x- and y-intercept formula for implicitly differentiated equations as follows.

If a function is defined implicitly, then its y-intercept formula, denoted as yyi, is given by ${{y}_{yi}}=y-x\frac{dy}{dx}$ and its x-intercept formula, denoted as yxi, is given by ${{y}_{xi}}=x-y\frac{dx}{dy}$.

The conic sections are great examples of implicit relationships. Here are examples for the three conics which are expressed as implicit equations.

Circles

The equation for circle is x² + y² = r² for radius of length r. Through implicit differentiation, the derivative is $\frac{dy}{dx}=-\frac{x}{y}$. So the y-intercept formula is ${{y}_{yi}}=y-x\left( \frac{-x}{y} \right)=\frac{{{x}^{2}}+{{y}^{2}}}{y}=\frac{{{r}^{2}}}{y}$, and x-intercept formula is ${{y}_{xi}}=x-y\left( -\frac{y}{x} \right)=\frac{{{x}^{2}}+{{y}^{2}}}{x}=\frac{{{r}^{2}}}{x}$.

Ellipses

The equation for an ellipse is $\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1$. The derivative is $\frac{dy}{dx}=-\frac{{{b}^{2}}x}{{{a}^{2}}y}$. The y-intercept formula is ${{y}_{yi}}=y+\frac{{{b}^{2}}{{x}^{2}}}{{{a}^{2}}y}=\frac{{{a}^{2}}{{y}^{2}}+{{b}^{2}}{{x}^{2}}}{{{a}^{2}}y}=\frac{{{a}^{2}}{{b}^{2}}}{{{a}^{2}}y}=\frac{{{b}^{2}}}{y}$, and the x-intercpet formula is ${{y}_{xi}}=\frac{{{a}^{2}}}{x}$.

Summary of Conics

The table below summarizes the slope, y-intercept, and x-intercept formulas.

 Conic Equation $\frac{dy}{dx}$ yyi yxi Circle x² + y² = r² $-\frac{x}{y}$ $\frac{r^2}{y}$ $\frac{r^2}{x}$ Parabola y = ax² 2ax -ax² x/2 Ellipse $\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1$ $-\frac{{{b}^{2}}x}{{{a}^{2}}y}$ $\frac{{{b}^{2}}}{y}$ $\frac{{{a}^{2}}}{x}$ Hyperbola $\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1$ $\frac{{{b}^{2}}x}{{{a}^{2}}y}$ $-\frac{{{b}^{2}}}{y}$ $\frac{{{a}^{2}}}{x}$

## Differential Formulas

We can find an expression for the y-intercept and x-intercept for a given function or relation. However, what if we wanted to do the reverse? For example, is there a function f(x) such that a tangent at (x, f(x)) passes through the x-intercept at (x – 1, 0)? Or is there a function f(x) such that a tangent at (x, f(x)) passes through the y-intercept at (0, ln x)? Techniques for solving differential equations can help us find the general formulas such that a tangent at (x, f(x)) passes though (P(x), 0) or (0, Q(x)).

The General Solution to the X-Intercept Differential Equation

The x-intercept formula is given by ${{y}_{xi}}=x-y\frac{dx}{dy}$. We let yxi = P(x) and solve for y in the x-intercept differential equation: $x-y\frac{dx}{dy}=P(x)$. We wish to find a function y for which its x-intercept is given by P(x).

(i) $x-y\frac{dx}{dy}=P(x)$

(ii) $\frac{x}{dx}-\frac{y}{dy}=\frac{P(x)}{dx}$ (Separate variables by dividing by dx.)

(iii) $\frac{x-P(x)}{dx}=\frac{y}{dy}$

(iv) $\frac{dy}{y}=\frac{dx}{x-P(x)}$

(v) $\int{\text{ }\frac{dy}{y}}=\int{\text{ }\frac{dx}{x-P(x)}}$ (Integrate both sides.)

(vi) $\ln \left| y \right|=\int{\text{ }\frac{dx}{x-P(x)}}+{{C}_{1}}$

(vii) $\left| y \right|={{e}^{\int{\frac{dx}{x-P(x)}+{{C}_{1}}}}}=C{{e}^{\int{\frac{dx}{x-P(x)}}}}$

If there exists a function y such that its x-intercept is given by the function P(x), then y is given by: $y=C{{e}^{\int{\frac{dx}{x-P(x)}}}}$

We disregard the absolute value bars since C can be positive or negative.

### Example 1

Problem: Find a function f such that its tangent at (x, f(x)) passes through (x², 0).

Solution: Using the formula, we must be able to integrate $\int{\text{ }\frac{dx}{x-{{x}^{2}}}}$ since P(x) = x². By means of partial fractions we have:

(i) $\int{\text{ }\frac{dx}{x-{{x}^{2}}}}=\int{\text{ }\frac{dx}{x(1-x)}}=\int{\text{ }\left( \frac{1}{x}+\frac{1}{(1-x)} \right)}\text{ }dx=\ln x-\ln (1-x)=\ln \frac{x}{1-x}$

Therefore, the function $y=C{{e}^{\ln [{x}/{(1-x)}\;]}}=\frac{Cx}{1-x}$ whose tangent at some point p has its x-intercept at (p², 0).

The General Solution to the Y-Intercept Differential Equation

For the general y-intercept formula, we must solve the y-intercept differential equation $y-x\frac{dy}{dx}=Q(x)$, where yyi = Q(x). The solution to a first-order linear differential equation $\frac{dy}{dx}+{{P}_{1}}(x)y={{Q}_{1}}(x)$ is given by: $y{{e}^{\int{{{P}_{1}}(x)\text{ }dx}}}=\int{\text{ }{{Q}_{1}}(x){{e}^{\int{{{P}_{1}}(x)\text{ }dx}}}\text{ }dx+C}$. The y-intercept differential equation can be manipulated to fit the first-order linear differential equation form.

(i) $y-x\frac{dy}{dx}=Q(x)\text{ }\Rightarrow \text{ }\frac{dy}{dx}-\frac{1}{x}y=-\frac{Q(x)}{x}$

Thus, we have ${{P}_{1}}(x)=-\frac{1}{x}$ and ${{Q}_{1}}(x)=-\frac{Q(x)}{x}$. Substitution in the solution formula for the first-order linear differential equation gives us:

(iii) $y{{e}^{-\int{1/x\text{ }dx}}}=-\int{\text{ }\frac{Q(x)}{x}{{e}^{-\int{1/x\text{ }dx}}}\text{ }dx}+C$

(iv) $y{{e}^{-\ln \left| x \right|}}=-\int{\text{ }\frac{Q(x)}{x}{{e}^{-\ln \left| x \right|}}\text{ }dx}+C$

(v) $y=-x\int{\text{ }\frac{Q(x)}{{{x}^{2}}}dx}+Cx$

We wish to make the formula less complicated, so we disregard the absolute value bars.

If there exists a function y such that its y-intercept is given by the function Q(x), then y is given by: $y=Cx-x\int{\text{ }\frac{Q(x)}{{{x}^{2}}}dx}$.

### Example 2

Problem: Find a function f(x) such that a tangent at (x, f(x)) passes through (0, ln x).

Solution: In this case Q(x) = ln x. Performing integration by parts, we have:

(i) $y=Cx-x\int{{{x}^{-2}}\ln xdx}=Cx-x\left[ -{{x}^{-1}}\ln x+\int{\text{ }{{x}^{-2}}\text{ }dx} \right]=-x\left( -\frac{\ln x}{x}-\frac{1}{x} \right)=Cx+1+\ln x$

The general solution is $Cx+1+\ln x$, for all real C.

In the above intercept differential equations, the x-intercept and y-intercept were functions of x. However, the intercepts do not have to be functions. If the x-intercept differential equation is y dependent, then the equation would be $\[x-y\frac{dx}{dy}=P(y)$, where P is a relation of y. In this equation, we cannot directly solve for y, but we can solve for x. We can manipulate this differential equation into the form $\frac{dx}{dy}-\frac{1}{y}x=-\frac{P(y)}{y}$. Using the solution to the first-order linear differential equation, we can solve for x to be $x=Cy-y\int{\text{ }\frac{P(y)}{{{y}^{2}}}dy}$.

If the x-intercept of a curve is given by P(y) and y dependent, then the curve is defined by the equation $x=Cy-y\int{\text{ }\frac{P(y)}{{{y}^{2}}}dy}$. The curve is a function only if we can isolate y.

## Last Word

A great application of the x- and y-intercept formulas is characterizing the tangent-generated curve. You can read about this topic here: tangent-generated curve.