# Quartic Function

A quartic function is a polynomial function of degree 4, meaning its highest power term is raised to the power of 4.

The general form of a quartic function is ax4 + bx3 + cx2 + dx + e, where a is any non-zero real number (a ≠ 0) and b, c, d, and e are any real numbers.

Here are some terms used to define a quartic function or a quartic graph.

1. As we know, quartic functions have a degree or exponent of four, thus different from the lower-degree polynomials like linear, quadratic, and cubic functions.
2. The leading coefficient ‘a’ influences the graph’s end behavior. If a is positive, the graph rises on both ends; if negative, it falls on both ends.
3. Quartic functions may exhibit symmetry. An even-powered term like x4 leads to symmetry about the y-axis, making the function even.
4. They can have up to three turning points, where the curve direction changes from increasing to decreasing or vice versa.

Here are a few examples of quartic functions:

• x4 – 1
• 6x4 – 5x3 + x – 9
• x4 – 6x + 12
• x4 + 2x2 + 1, called a biquadratic function (a special form of a quartic function, where the coefficients of x3 and x are zero).

Quartic functions have derivatives of 1 to 3 roots that can be solved by the rational root theorem and factoring, while the biquadratic functions are solved using the quadratic formula.

## Finding Roots

The roots of a quartic function are the values of x that make the function equal to 0. It is done using the rational root theorem and factoring.

The rational root theorem for a quartic function ax4 + bx3 + cx2 + dx + e states that if the function has any rational roots, these roots could be formed by dividing some combinations of factors of the last term (constant term) by factors of the first term’s coefficient (the coefficient of x4.)

Let us find the roots of a quartic function x4 + 2x3 – 7x2 – 8x + 12

Since factors of 12 are 1, 2, 3, 4, 6, and 12. The possible solutions are in the ratio: ${\pm \dfrac{1}{1}}$, ${\pm \dfrac{2}{1}}$, ${\pm \dfrac{3}{1}}$, ${\pm \dfrac{4}{1}}$, ${\pm \dfrac{6}{1}}$, and ${\pm \dfrac{12}{1}}$.

Now, taking any of these solutions to substitute in the above function to verify if the equation is 0, we get

(1)4 + 2(1)3 – 7(1)2 – 8(1) + 12 = 1 + 2 – 7 – 8 + 12 = 0

Thus, 1 is the root of the given quartic equation, and one of its factors is (x – 1).

Dividing the given quartic function by (x – 1), we get

x4 + 2x3 – 7x2 – 8x + 12

= (x – 1)(x3 + 3x2 – 4x – 12)

Now, we further factor the polynomial x3 + 3x2 – 4x – 12 to obtain all the roots.

Here, x3 + 3x2 – 4x – 12

= x2(x + 3) – 4(x + 3)

= (x + 3)(x2 – 4)

= (x + 3)(x + 2)(x – 2)

Now, the quartic function can be written in the factored form as x4 + 2x3 – 7x2 – 8x + 12 = (x – 1)(x – 2)(x + 2)(x + 3)

Thus, the roots of the given quartic equation x4 + 2x3 – 7x2 – 8x + 12 = 0 are -3, -2, 1, and 2.

## How to Graph a Quartic Function

Graphing a quartic function is important to analyze its behavior. When graphed, quartic functions often create a shape with two humps or two troughs, reflecting the influence of the x4 term. The number and nature of turning points depend on the coefficients in the function.

Let us graph the quartic function y = f(x) = x4 – 5x2 + 4

Step 1: By putting x = 0, we find the y-intercept of the function and then solve for y = f(x), we get

f(0) = (0)4 – 5(0)2 + 4 = 4

The y-intercept is (0, 4)

Step 2: Now, we find the x-intercepts of the function by factoring.

Considering p = x2, we get

f(p) = p2 – 5p + 4

= p(p – 4) + 1(p – 4)

= (p – 4)(p – 1)

Now, setting f(p) = 0, we get

(p – 4) = 0 and (p – 1) = 0

⇒ p = 4 and 1.

Placing the values of p

⇒ x2 = 4 and x2 = 1

⇒ x = ±2 and x = ±1

Thus, the x-intercepts of the quartic parent function are (2, 0), (-2, 0), (1, 0), and (-1, 0).

Step 3: Now, selecting some points between each of the x-intercepts and evaluating the values of the function, we get

Step 4: Plotting some points between each of the x and y intercepts from step 3 and connecting the points with a curve, we get

Here, in the above graph, the curve has three turning points. We observe that the leading coefficient of the function is positive, and then both ends of the curve go up to infinity.

When the leading coefficient of the function is negative, then both ends of the curve go downward to infinity.

## Domain and Range

Since all possible input values are the domain and all possible output values are the range of a function, the domain of the above quartic equation is (-∞, ∞) (or -∞ < x < ∞, {x:x Є ℝ}), and its range is [-2.25, ∞) (or f(x) ≥ -2.25).

## Solved Examples

Example 1: Find the roots of the quartic equation x4 – 2x3 – 3x2 + 4x + 4 = 0

Since factors of 4 are 1, 2, and 4. The possible solutions are in the ratio: ${\pm \dfrac{1}{1}}$, ${\pm \dfrac{2}{1}}$, and ${\pm \dfrac{4}{1}}$.

Now, taking any of these solutions to substitute in the above function to verify if the equation gives the value 0, we get

(-1)4 – 2(-1)3 – 3(-1)2 + 4(-1) + 4 = 1 + 2 – 3 – 4 + 4 = 0

Thus, -1 is the root of the given quartic equation, and one of its factors is (x + 1).

Dividing the given quartic function by (x + 1), we get

x4 – 2x3 – 3x2 + 4x + 4

= (x + 1)(x3 – 3x2 + 4)

Similarly, we further factor the polynomial x3 – 3x2 + 4 to obtain all the roots.

Since, for x = -1, the equation equals 0

Thus we get another factor as (x + 1).

Dividing the given cubic function by (x + 1), we get

x3 – 3x2 + 4

= x2(x + 1) – 4x(x + 1) + 4(x + 1)

= (x + 1)(x2 – 4x + 4)

Further simplifying the quadratic function x2 – 4x + 4, we get

x2 – 4x + 4

= (x – 2)2 = (x – 2)(x – 2)

Now, the quartic function can be written in the factored form as x4 – 2x3 – 3x2 + 4x + 4 = (x + 1)(x + 1)(x – 2)(x – 2)

Thus, the roots of the given quartic equation x4 – 2x3 – 3x2 + 4x + 4 = 0 are -1, -1, 2, and 2.

Example 2: Plot the graph of a quartic function: x4 + 4x3 + 3x2 – 4x – 4

By putting x = 0, we find the y-intercept of the function and then solve for the value of y to get

(0)4 + 4(0)3 + 3(0)2 – 4(0) – 4 = -4

The y-intercept is (0, -4)

Now, we find the x-intercepts of the function by factoring.

x4 + 4x3 + 3x2 – 4x – 4

= x3(x + 1) + 3x2(x + 1) – 4(x + 1)

= (x + 1)(x3 + 3x2 – 4)

= (x + 1){x2(x – 1) + 4x(x – 1) + 4(x – 1)}

= (x + 1)(x – 1)(x2 + 4x + 4)

= (x – 1)(x + 1)(x + 2)(x + 2)

Setting f(x) = 0, we get

(x – 1)(x + 1)(x + 2)(x + 2) = 0

⇒ x = 1, -1, -2, and -2

The x-intercepts of the function are (1, 0), (-1, 0), (-2, 0), and (-2, 0).

Now, selecting some points between each of the x-intercepts and evaluating the values of the function, we get

By plotting some points between each of the x and y intercepts from step 3 and connecting the points with a curve, we get the quartic graph.

Example 3: Which of the following graphs represents a quartic function?

Here, option c) represents a quartic function.