Warm-up solution

The graph of the function \(y=\dfrac{1}{x^2}\) is stretched as shown in this diagram.

If the graph is stretched by a factor of \(b\) in the \(x\)-direction (as shown), what is the equation of the resulting graph?

One way to think about this problem is to use the language of functions.

The original graph has equation \(y=f(x)\), where \(f(x)=\dfrac{1}{x^2}\).

When we stretch the graph by a factor of \(b\) in the \(x\)-direction, a point with coordinates \((x,f(x))\) is stretched to the point with coordinates \((bx,f(x))\), as shown in the following diagram. If we write \(X=bx\) for the new \(x\)-coordinate, then the new point has coordinates \((X,Y)=(X,f(X/b))\), so the new graph has equation \(Y=f(X/b)\), or \(y=f(x/b)\) using lower-case letters.

A graph of y equals 1 over x squared, then shown stretched by a scale factor b in the x direction with the point x comma f of x shown

So the equation of the stretched graph is \(y=f(x/b)=\dfrac{1}{(x/b)^2}\). Multiplying the numerator and denominator by \(b^2\) gives the simpler form \(y=\dfrac{b^2}{x^2}\).

And if, instead, the original graph is stretched by a factor of \(c\) in the \(y\)-direction, what is the equation of the resulting graph?

This time, the point \((x,f(x))\) is stretched to the point with coordinates \((x,cf(x))\), as shown in the following diagram. Writing the coordinates of the new point again as \((X,Y)=(x,cf(x))\), we find the new graph has equation \(Y=cf(X)\), as \(X=x\), or \(y=cf(x)\) using lower-case letters.

A graph of y equals 1 over x squared, then shown stretched by a scale factor c in the y direction with the point x comma f of x shown

Therefore the equation of the stretched graph is \(y=\dfrac{c}{x^2}\).