# The X- and Y-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 \(f_{xi}(x)=-\frac{f_{yi}(x)}{f'(x)}\).

** 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 \(f_{xi}(x)= -\frac{f_{yi}(x)}{f'(x)} = x-\frac{f(x)}{f'(x)}\), 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).

The image below show the representation of the designations.

## 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 \(\frac{x\cdot f'(x)-f(x)}{f'(x)}\). 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 \(r_1 \approx x-\frac{f(x)}{f'(x)}=\) \(x-\frac{x^2-4}{2x}=\) \(\frac{x^2+4}{2x}\). 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 | \(x^2+y^2=r^2\) | \(-\frac{x}{y}\) | \(\frac{r^2}{y}\) | \(\frac{r^2}{x}\) |

Parabola | \(y = ax^2\) | \(2ax\) | \(-ax^2\) | \(x/2\) |

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 | \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) | \(\frac{b^2 x}{a^2 y}\) | \(-\frac{b^2}{y}\) | \(\frac{a^2}{x}\) |

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

(vi) \(\ln \left| y \right| = \int \frac{dx}{x-P(x)}+C_1\)

(vii) \(\left| y \right| = e^{\int \frac{dx}{x-P(x)} + C_1} = Ce^{\int \frac{dx}{x-P(x)}}\)

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

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 \(\int \frac{dx}{x-x^2}\) since *P*(*x*) = *x*². By means of **partial fractions** we have:

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

Therefore, the function \(y=Ce^{\ln [{x}/{(1-x)}]} = \frac{Cx}{1-x}\) whose tangent at some point *p* has its *x*-intercept at (*p*², 0).

For the example above, we have graphed the function \(y = \frac{2x}{1-x}\), undoubtedly a hyperbola. Two tangents are shown in the two branches of the hyperbola. They both cross the *x*-axis at a point that is squared of the *x* value of the tangency. (Values are approximated.)

### 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: \(ye^{\int P_1(x)dx} = \int Q_1(x)e^{\int P_1(x)dx}dx+C\). The

*y*-intercept differential equation can be manipulated to fit the first-order linear differential equation form.

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

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

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

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

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

The function *y* such that its *y*-intercept is given by the function *Q*(*x*), then *y* is given by: \(y=Cx-x\int{\text{ }\frac{Q(x)}{{{x}^{2}}}dx}\). This function is graphed below showing two points of tangency and their *x*-intercepts.

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

The general solution is \(Cx+1+\ln x\), for all real *C*. This function is graphed on the right. The green function is when *C* = 2: \(f(x) = 1+2x+\ln(x)\). The blue function is when *C* = –3: \(f(x) = 1 - 3x + \ln(x)\). The *y*-intercept is independent of *C* and the tangent always crosses the *y*-axis at ln(*x*).

If you were to track the tangent points on this graph and place them at *x* = 1, then the *y*-intercepts will be 0 because ln(1) = 0. Because *C* = –3 in the blue graph, it has a relative mamixum, which can be found at *x* = 1/3. At this point, since the tangent is horizontal, it follows that *f*(1/3) = *f _{yi}*(1/3) ≈ –1.098.

If *C* is positive, the derivative has no zeroes. Hence, there are no relative extrema.

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 \(x = Cy-y\int \frac{P(y)}{y^2}dy\).

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

## Constant Ratio Curves

An interesting excursion with the intercept formulas are the constant ratio curves given by the equation \(y=\frac{C}{x^{\frac{1}{k}}}\). Depicted below is the curve for *k* = 4 and *C* = 1.

In this family of curves, the ratio AB to AD is always constant for a point B that is tangent to the curve at A. This ratio is equal to 1/*k*. The constant *C* plays no role in the ratio. Therefore, for all *C*, the curve \(y=\frac{C}{x^{\frac{1}{5}}}\) will produce a ration of 1/5.

When *k* = 1, we have a family of hyperbolas. The tangents of the hyperbola divides the segment into equal parts.

Above image shows the equation \(y=\frac{0.5}{x^2}\) graphed. It shows that the tangent is divided into two at the point of tangency.

Even if *k* is less than 1, the constant ratio holds true. As a matter of fact, when the ratio is less than 0, the equation becomes a simple polynomial. If the ratio is –1/2, the equation is \(y = \frac{1}{2}x^2\) which is the familiar parabola. In this case where the point of tangency is not in the middle of the *x*- and *y*-intercepts, the ratio we measure is from the point of tangency to the *y*-axis with the point of tangency to the *x*-axis. The segment itself is dissected into half by the axis, as can be seen below.

### Example 3

In general, a tangent to the simple polynomial \(y = x^n\) is divided in a ratio of 1 to (*n* – 1) by the *x*-intercept. Let’s prove this. Let point P be given by (*p*, *p*^{n}) on the curve. The slope of the tangent at this point is:

(i) \(y'(p) = np^{n-1}\)

The *y*-intercpet is given by:

(ii) \(y_{yi}(p) = p^n - p(np^{n-1}) = (1-n)p^n\).

So the *y*-intercept is located at \((0, (1-n)p^n)\). The *x*-intercept is given by:

(iii) \(y_{xi}(p) = p - p^n\cdot\frac{1}{np^{n-1}} = p-\frac{1}{n}p = \frac{n-1}{n}p\).

So the *x*-intercept is located at \(\left(\frac{n-1}{n}p, 0\right)\). Now, we find the distances between the tangent point with the intercept points. First, the distance, *d _{y}*, between the tangent point (

*p*,

*p*

^{n}) and

*y*-intercept \((0, (1-n)p^n)\):

(iv) \(d_y = \sqrt{(p - 0)^2 + [p^n - (1-n)p^n]^2}\)

(v) \(d_y = \sqrt{p^2 + n^2p^{2n}}\)

(vi) \(d_y = p\sqrt{1 + n^2p^{2n-1}}\)

Now, we will find the distance between the tangent point (*p*, *p*^{n}) and *x*-intercept \(\left(\frac{n-1}{n}p, 0\right)\):

(vii) \(d_x = \sqrt{\left(p - \frac{n-1}{n}p\right)^2 + (p^n - 0)^2}\)

(viii) \(d_x = \sqrt{\left(\frac{p}{n}\right)^2 + p^{2n}}\)

(xv) \(d_x = \frac{p}{n}\sqrt{1 + n^2p^{2n-2}}\)

When we take the ratio of the distances, we get \(\frac{d_y}{d_x} = \frac{p\sqrt{1 + n^2p^{2n-1}}}{\frac{p}{n}\sqrt{1 + n^2p^{2n-2}}} = n\). This is exactly what we expected. The ratio between the distances is *n*. And because the point of tangency is not between the intercepts, the segment from the point of tangency to the *y*-intercept is divided in a ratio of 1 : (*n* – 1) by the *x*-intercept.

## 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.