Quartic And Other Polynomial Functions
Understand quartic and higher polynomial functions for Queensland Year 11 Mathematical Methods (QCAA). A quartic has degree four, and its shape depends on the degree and leading coefficient.
You will learn to predict the behaviour for large positive and negative x, sketch a transformed quartic with a turning point, read multiplicity at each zero, and solve a quartic that behaves like a quadratic — supporting skills that extend the cubic work.
Every question with a fully worked solution.
- Quartic And Other Polynomial Functions - Video - Quartic functions Watch
Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), a quartic is a degree-4 polynomial. Because the degree is even, both ends point the same way — up when \(a>0\), down when \(a<0\). This page shows how to read the degree and leading coefficient, the turning point of \(y=a(x-h)^4+k\), the intercepts and multiplicity of a factorised quartic, and how to solve a quadratic-in-\(x^2\) quartic — all without calculus.
A quartic has the form \(y=ax^4+bx^3+cx^2+dx+e\) with \(a\neq0\). Its degree is \(4\) and its leading coefficient is \(a\). Because \(4\) is even, the two ends behave the same way: both up if \(a>0\), both down if \(a<0\).
The turning-point form \(y=a(x-h)^4+k\) has a single turning point at \((h,k)\): a minimum when \(a>0\) and a maximum when \(a<0\). The value \(k\) is then the smallest (or largest) value of \(y\).
A factorised quartic shows its zeros. As with cubics, an even multiplicity (a squared factor) makes the graph touch the axis, and an odd multiplicity makes it cross.
End behaviour of a quartic (even degree):
Turning-point form:
Solving a quadratic-in-\(x^2\) by letting \(u=x^2\):
How to analyse and sketch a quartic
- Degree and \(a\): confirm degree \(4\) and read the leading coefficient; even degree means both ends match the sign of \(a\).
- Intercepts: find the \(y\)-intercept at \(x=0\) and the \(x\)-intercepts from the factors or by solving.
- Multiplicity: mark each \(x\)-intercept as a touch (even power) or a cross (odd power).
- Join up: use the matching ends and the turning point (for the \((x-h)^4+k\) form) to draw a smooth curve.
Degree and leading coefficient — read the highest power:
| \(\text{degree}\) | \(=\) | \(4\) |
| \(\text{leading coefficient }a\) | \(=\) | \(-2\) |
The degree is even and \(a=-2<0\), so both ends point the same way — down.
End behaviour:
| \(\text{left}\) | \(\to\) | \(-\infty\) |
| \(\text{right}\) | \(\to\) | \(-\infty\) |
y-intercept — substitute \(x=0\):
| \(y\) | \(=\) | \(-2(0)^4+3(0)^2-1\) |
| \(=\) | \(-1\) |
Degree \(4\), \(a=-2\); both ends \(\to-\infty\); \(y\)-intercept \(-1\).
Turning point — compare with \(y=a(x-h)^4+k\), where \(a=1>0\) (a minimum):
| \((h,k)\) | \(=\) | \((1,-16)\) |
y-intercept — substitute \(x=0\):
| \(y\) | \(=\) | \((0-1)^4-16\) |
| \(=\) | \(1-16\) | |
| \(=\) | \(-15\) |
x-intercepts — set \(y=0\) and solve:
| \((x-1)^4-16\) | \(=\) | \(0\) |
| \((x-1)^4\) | \(=\) | \(16\) |
| \((x-1)^2\) | \(=\) | \(4\) |
Take the positive root only (a square cannot be \(-4\)):
| \(x-1\) | \(=\) | \(\pm 2\) |
| \(x\) | \(=\) | \(3\ \text{or}\ -1\) |
Minimum \((1,-16)\); \(x\)-intercepts \(-1\) and \(3\); \(y\)-intercept \(-15\).
x-intercepts — set each factor to \(0\):
| \((x+2)(x-1)^2(x-3)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(-2,\ 1,\ 3\) |
At \(x=1\) the squared factor gives a touch; at \(x=-2\) and \(x=3\) the graph crosses.
y-intercept — substitute \(x=0\):
| \(y\) | \(=\) | \((0+2)(0-1)^2(0-3)\) |
| \(=\) | \((2)(1)(-3)\) | |
| \(=\) | \(-6\) |
End behaviour — expanding gives leading term \(x^4\), so \(a=1>0\):
| \(\text{both ends}\) | \(\to\) | \(+\infty\) |
Crosses at \(-2,3\), touches at \(1\), \(y\)-intercept \(-6\); both ends \(\to+\infty\).
Substitute \(u=x^2\):
| \(u^2-13u+36\) | \(=\) | \(0\) |
Factorise the quadratic in \(u\):
| \((u-4)(u-9)\) | \(=\) | \(0\) |
| \(u\) | \(=\) | \(4\ \text{or}\ 9\) |
Replace \(u=x^2\) — both values are positive, so all give real \(x\):
| \(x^2\) | \(=\) | \(4\ \Rightarrow\ x=\pm 2\) |
| \(x^2\) | \(=\) | \(9\ \Rightarrow\ x=\pm 3\) |
Solutions \(x=-3,-2,2,3\); the same four \(x\)-intercepts.
Common pitfalls
Frequently asked questions
What is the end behaviour of a quartic?
Because the degree \(4\) is even, both ends go the same way: both up if \(a>0\), both down if \(a<0\).
How do you find the turning point of y = a(x-h)^4 + k?
It is at \((h,k)\): a minimum when \(a>0\) and a maximum when \(a<0\), with \(k\) the least or greatest value.
How do you solve a quartic like x^4 - 13x^2 + 36 = 0?
Let \(u=x^2\) to get \(u^2-13u+36=0\), solve for \(u\), then take \(x=\pm\sqrt{u}\), rejecting any negative \(u\).
How many x-intercepts can a quartic have?
A quartic can have \(0,1,2,3\) or \(4\) real \(x\)-intercepts, depending on how the factors and their multiplicities line up.
What does a squared factor do to a quartic graph?
It makes the graph touch the \(x\)-axis and turn back, rather than crossing through it.
Do you need calculus for quartics in Year 11?
No. You sketch from the degree, leading coefficient, intercepts, multiplicity and the \((x-h)^4+k\) turning point — not from derivatives.