Problem of the Day


Let (p,q) be a point on the graph of y=x^n where p \neq 0. If the value of n is chosen randomly from the set \{-1,0,1,2,3,4,5\}, what is the probability that point (-p,-q) is also on the graph?

A. \displaystyle \frac{1}{7}

B. \displaystyle\frac{3}{7}

C. \displaystyle\frac{4}{7}

D. \displaystyle\frac{5}{7}

E. \displaystyle\frac{6}{7}

Reveal Answer

Answer

[latexpage]

C. $\displaystyle\frac{4}{7}$

See the Solution

Solution

[latexpage]

We are essentially trying to determine the values of $n$ in the set ${-1,0,1,2,3,4,5}$ such that (p,q) being on the graph of $y=x^n$ implies that (-p,-q) is also on the graph. Functions that have this property are called odd functions, and their graphs exhibit symmetry about the origin in the coordinate plane. This means that if you rotate the graph $180^o$ about the origin, you will produce the exact graph you started with. You don’t necessarily need to be familiar with the graphs of the different functions of the form $y=x^n$ to solve this problem, but it might help.

The function $y=x^n$ is odd if and only if $n$ is odd. For example, take $n=3$. If $y=x^3$, we know the point $(2,8)$ is on the graph because it satisfies the equation. $(-2,-8)$ would also be on the graph because $(-2)^3=-8$. Every point that appears on the graph of $y=x^3$ would have this same property. That is, plugging in the opposite $x$-coordinate of some point on the graph gives you the opposite $y$-coordinate of that point. Every odd value of $n$ produces a function $y=x^n$ that has this property.

If $n$ is even, the function $y=x^n$ no longer has this property. For example, if $n=2$ and $y=x^2$, plugging in the opposite $x$ value gives you the same $y$ value. Functions with this property are called even functions, and their graphs are symmetrical about the y-axis. These functions would have the property that $(p,q)$ on the graph implies $(-p,q)$ is on the graph, but not $(-p,-q)$. The only exception would be the point $(0,0)$, but we don’t have to consider that in this problem because we are told that $p \neq 0$.

So, 4 out of the 7 possible values of $n$ produce an odd function, so the probability we are looking for is $\displaystyle \frac{4}{7}$

If you’d like to see the symmetry of the graphs that I’m talking about, go to www.graph.tk and graph $y=x^n$ for the different values of $n$


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