# The Intercept Formulas and Newton's Method

# 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 *f _{yi}*(

*x*) or

*f*or

_{yi}*y*. The

_{yi}*x*-intercept formula is denoted

*f*(

_{xi}*x*) or

*f*or

_{xi}*y*. So a tangent to

_{xi}*f*(

*x*) at (

*p*,

*f*(

*p*)) crosses the

*y*-axis at (0,

*f*(

_{yi}*p*)) and

*x*-axis at (

*f*(

_{xi}*p*), 0). One application of the

*x*-intercept formula is

**Newton's Method**for finding the real zeros of a polynomial.

## Derivation of the Formulas

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 *f _{yi}*(

*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

*f*(

_{yi}*x*) gives the

*y*value of the point where a tangent to the curve at (

*x*,

*f*(

*x*)) crosses the

*y*-axis or (0,

*f*(

_{yi}*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

*f*(

_{xi}*x*) gives the

*y*value of the point where a tangent to the curve at (

*x*,

*f*(

*x*)) crosses the

*x*-axis or (

*f*(

_{xi}*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 *r*_{1} 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:

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 *y _{yi}*, is given by ${{y}_{yi}}=y-x\frac{dy}{dx}$ and its

*x*-intercept formula, denoted as

*y*, is given by ${{y}_{xi}}=x-y\frac{dx}{dy}$.

_{xi}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 *x*² + *y*² = *r*² 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 | $\frac{dy}{dx}$ | y_{yi} | y_{xi} |

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 *y _{xi}* =

*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 (*x*², 0).

Solution: Using the formula, we must be able to integrate $\int{\text{ }\frac{dx}{x-{{x}^{2}}}}$ since *P*(*x*) = *x*². 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 (*p*², 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 *y _{yi}* =

*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.