# What are the x- and y-intercepts, if any, of the function? f(x) = 9x^2\ln{4x}

## Question:

What are the x- and y-intercepts, if any, of the function?

{eq}f(x) = 9x^2\ln{4x} {/eq}

## Intercepts of a Function:

The {eq}x {/eq}-intercepts of a graph are the intersection points of that graph with the {eq}x {/eq}-axis, and the {eq}y {/eq}-intercepts of a graph are the intersection points of that graph with the {eq}y {/eq}-axis. The intercepts of a function {eq}f(x) {/eq} are defined to be the intercepts of the graph {eq}y=f(x) {/eq}, which means that:

1) The {eq}x {/eq}-intercepts of {eq}f(x) {/eq} are the points {eq}(c,0) {/eq}, where {eq}f(c)=0 {/eq} (i.e., {eq}c {/eq} is a root of {eq}f {/eq}), and

2) The {eq}y {/eq}-intercept of {eq}f(x) {/eq} is the point {eq}(0,f(0)) {/eq}, if {eq}0 {/eq} is in the domain of {eq}f {/eq}.

The function {eq}f(x)=9x^2\ln(4x) {/eq} is only well-defined for {eq}x>0 {/eq}, because {eq}\ln u {/eq} is only well-defined for {eq}u>0 {/eq}. This immediately implies that {eq}0 {/eq} is not in the domain of {eq}f(x) {/eq}, so {eq}f(x) {/eq} has no {eq}y {/eq}-intercept.

To find the {eq}x {/eq}-intercepts of {eq}f {/eq}, we need to find the roots of {eq}f {/eq}. The first factor {eq}9x^2 {/eq} is never zero for {eq}x>0 {/eq}, so {eq}f(x)=0 {/eq} when the second factor is zero; that is, when:

{eq}\begin{align*} \ln(4x)&=0\\ 4x&=1&&\text{(exponentiating both sides)}\\ x&=\frac{1}{4} \, . \end{align*} {/eq}

In summary, {eq}f(x) {/eq} has a single {eq}x {/eq}-intercept at {eq}\left(\frac{1}{4},0\right) {/eq}, and no {eq}y {/eq}-intercepts.