9. A square has vertices at $(0,0),(1,0),(1,1)$ and $(0,1)$. Graphs of the following equations are drawn on the same set of axes as the square.
$$
x^{2}+y^{2}=1, \quad y=x+1, \quad y=-x^{2}+1, \quad y=x, \quad y=\frac{1}{x}
$$

How many of the graphs pass through exactly two of the vertices of the square?
A 1
B 2
C 3
D 4
E 5
SolUtion
C
Method 1
One method is to draw the graphs.
The graph of the equation $x^{2}+y^{2}=1$ is the circle with centre $(0,0)$ and radius 1. [By Pythagoras' Theorem, the distance, $d$ of the point with coordinates $(x, y)$ from the point $(0,0)$ is given by the equation $x^{2}+y^{2}=d^{2}$. Therefore the equation $x^{2}+y^{2}=1$ is satisfied by those points whose distance from the origin, $(0,0)$ is 1 . These are the points on the circle with centre $(0,0)$ and radius 1 , as shown in the diagram below.]

The graph of $y=x+1$ is the straight line with slope 1 that meets the $y$-axis at the point $(0,1)$.
The graph of $y=-x^{2}+1$ is a $\cap$-shaped parabola which is symmetrical about the $y$-axis and goes through the point $(0,1)$.

The graph of $y=x$ is the straight line with gradient 1 which goes through the point $(0,0)$.
The graph of $y=\frac{1}{x}$ is a rectangular hyperbola with the $x$-axis and $y$-axis as its asymptotes. In the diagram below we have just shown part of one arm of this hyperbola.
These graphs are shown in the diagram below.
From these graphs we see that the graphs of $x^{2}+y^{2}=1, y=-x^{2}+1$ and $y=x$ pass through two vertices of the square, whereas each of the other graphs passes through just one vertex of the square.
Hence the number of graphs that pass through exactly two vertices of the square is 3 .
MetHod 2
We check whether a given graph passes through a particular vertex of the square, by seeing whether the coordinates of the vertex satisfy the equation of the graph.
For example, when $x=0$ and $y=0$ we have $x^{2}+y^{2} \neq 1$. Therefore the graph of $x^{2}+y^{2}=1$ does not go through the vertex $(0,0)$.
On the other hand, when $x=0$ and $y=1$, we have $x^{2}+y^{2}=1$. Therefore the graph of $x^{2}+y^{2}=1$ does go through the vertex $(0,1)$.
In the table below we have shown the outcome of all these calculations. We have put 'yes' if the graph goes through the vertex in question, and 'no' if it does not.
\begin{tabular}{cccccc}
\hline vertices & $x^{2}+y^{2}=1$ & $y=x+1$ & $y=-x^{2}+1$ & $y=x$ & $y=\frac{1}{x}$ \\
\hline$(0,0)$ & no & no & no & yes & no \\
$(1,0)$ & yes & no & yes & no & no \\
$(1,1)$ & no & no & no & yes & yes \\
$(0,1)$ & yes & yes & yes & no & no \\
\hline
\end{tabular}

We see that in the case of three of the graphs considered, there are exactly two occurrences of 'yes' in the relevant column. We deduce that the number of graphs that pass through exactly two vertices of the square is 3 .