Where y = x⁴/(x + 1) Meets y = x²

Grades 10-11

Question: y = x^4/(x+1) y = x^2

To find the relationship and intersection points between the two functions

y = \frac{x^4}{x+1}

and

y = x^2

, we look for the values of

x

where their

y

-coordinates are equal.

We set the two equations equal to each other:

\begin{align*} \frac{x^4}{x+1} &= x^2 \\ x^4 &= x^2(x + 1) \quad (\text{assuming } x \neq -1) \\ x^4 &= x^3 + x^2 \\ x^4 - x^3 - x^2 &= 0 \end{align*}

Factoring out the common term

x^2

\begin{align*} x^2(x^2 - x - 1) = 0 \end{align*}

This gives us two cases for

x

Case 1: $x^2 = 0 \implies x = 0$ .
Case 2: $x^2 - x - 1 = 0$ . Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ : $\begin{align*} x &= \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} \\ x &= \frac{1 \pm \sqrt{5}}{2} \end{align*}$

The values for

x

are approximately

0

1.618

(the Golden Ratio,

\phi

), and

-0.618

(the conjugate,

\psi

By plugging these

x

-values into

y = x^2

, we find the corresponding points:

Point A (0, 0): Since $x^2$ was a squared factor in our equation, the two curves are tangent at the origin. Both curves have a horizontal slope ( $y' = 0$ ) at this point.
Point B (1.618, 2.618): The curves cross here.
Point C (-0.618, 0.382): The curves cross here.

Vertical Asymptote: The function $y = \frac{x^4}{x+1}$ has a vertical asymptote at $x = -1$ because the denominator becomes zero. To the left of $x = -1$ , the function tends toward $-\infty$ , while to the right, it tends toward $+\infty$ .
End Behavior: As $|x|$ becomes very large, $\frac{x^4}{x+1}$ behaves similarly to $x^3$ . This means it grows much faster than the parabola $y = x^2$ for positive $x$ .
Visual Guide: In the diagram, the blue curve represents the rational function, and the red curve represents the parabola. Note how they "kiss" at the origin and intersect at two other specific locations.