\(\blacktriangleright \) Graphically, shifting the domain of a function appears to shift the graph horizontally - left or right. The shape of the graph doesn’t change. The whole graph moves rigidly left or right. All points move the same distance horizontally.

• Let \(F(x)\) be a function with its domain.
• Let \(G(t)\) be a new function defined as \(G(t) = F(t+d_0)\) with its induced domain, where \(d_0\) is a fixed constant.

We begin with the function \(F\). The domain values of \(F\) are represented by \(x\). We then define a new function, \(G\). The domain values of \(G\) are represented by \(t\).

\(F\) and \(G\) are connected. To evaluate \(G\) at \(t\), you evaluate \(F\) at \(t + d_0\), where \(d_0\) is some constant. Therefore, \(t + d_0\) are \(x\)-values. We have \(x = t + d_0\). This tells us what to do to \(t\) to get \(x\). However, that is not how our story is told. Our story begins with \(F\) and then \(G\) is defined from \(F\).

We want to know how to get \(t\) from \(x\).

• \(x = t + d_0\)
• \(x - d_0 = t\)

To get corresponding values of \(t\) from values of \(x\), subtract \(d_0\).

\(\blacktriangleright \) The graph of \(G\) is obtained from the graph of \(F\) by subtracting \(d_0\) from domain values of \(F\), which appears to be reverse of the definition, \(G(t) = F(t+d_0)\). That is because the definition tells how to get the old domain for \(F\), rather than getting the new domain for \(G\).

Shifting Horizontally

Graph of \(y = T(v)\).

Define a new function \(W\) by \(W(h)=T(h+3)\), with the induced domain.

Which graph below is the graph of \(z=W(h)\)?

graph on the left graph on the right

To figure this out, we need to know how to get the new variable \(h\) from the old variable \(v\).

• \(h\) represents the domain of \(W\).
• \(v\) represents the domain of \(T\).
• \(v\) = \(h+3\)

These tell us that \(v-3 = h\) and the graph of \(W\) is the graph of \(T\) shifted left \(3\).

The shape of the graph didn’t change. It just slid to the left. There are still three pieces.

• The domain still has a gap of length \(2\) in it.
• The middle piece is still horizontal.
• The solid and hollow endpoint dots are still in the same positions.

Shifting doesn’t change the shape of the graph or any of the relative measurements. It is a rigid movement.

Sine

Graph of \(y = \sin (\theta )\).

Let’s create a new function called \(T\), which is a shifted sine function.

\[ T(t) = \sin (t + \tfrac {\pi }{6}) \]

This graph looks the same as the one above, except shifted to the left.

\(\theta = t + \tfrac {\pi }{6}\) or \(\theta - \tfrac {\pi }{6}= t \)

Labelling the horizontal axes \(\theta \) and \(t\) makes it easier to compare.

Sine

Graph of \(y = \sin (\theta )\).

\(\sin (\theta )\) is periodic with a period of \(2\pi \). If it is shifted by \(2\pi \) or any integer multiple of \(2\pi \), then the resulting function is again \(\sin (\theta )\).

\[ \sin (\theta + 2k\pi ) = \sin (\theta ) \, \text { where } \, k \in \textbf {Z} \]

• The zeros of \(\sin (\theta )\) are all integer multiples of \(\pi \).
• The maximum value is \(1\) and it occurs at: \(\left \{ \frac {\pi }{2} + 2k\pi \, | \, k \in \textbf {Z} \right \} = \left \{ \frac {(4k+1)\pi }{2} \, | \, k \in \textbf {Z} \right \}\)
• The minimum value is \(-1\) and it occurs at: \(\left \{ \frac {3\pi }{2} + 2k\pi \, | \, k \in \textbf {Z} \right \} = \left \{ \frac {(4k+3)\pi }{2} \, | \, k \in \textbf {Z} \right \}\)

If \(\sin (\theta )\) is shifted left by \(\frac {\pi }{2}\), then we get \(\cos (\theta )\).

\[ \sin \left (\theta + \frac {\pi }{2}\right ) = \cos (\theta ) \]

Graph of \(y = \cos (\theta )\).

This agrees with the unit circle. As you move along the unit circle, the right/vertical coordinate (sine) has the same value as the left/horizontal coordinate (cosine) back a quarter-circle.

Absolute Value

Graph of \(y = |x|\).

What does the graph of \(z = |t-2|\) look like?

This is a shift from the basic absolute value graph. All we really need to know is where is the corner.

The corner occurs when the inside of the absolute value equals \(0\).

\(t-2=0\) when \(t=2\). That is where the new corner sits.

Much of graphing follows this example.

There are important/strategic points for the function’s graph. You identify the new position of those points. Then, the shifted graph follows the basic shape of the original graph.

For example, \(y = \log _k(t)\) [ including \(y = \ln (t)\) ] has a vertical asymptote when the inside of the logarithm equals \(0\). The zero, and corresponding horizontal intercept, occurs when the inside equals \(1\).

Here is the graph of \(y = \log _2(t+5)\).

• vertical asymptote: \(t+5=0\), when \(t=\answer {-5}\)
• horizontal intercept: \(t+5=1\), when \(t=\answer {-4}\)

The graph looks the same as the basic logarithm graph, just slid left \(5\).

Shifting Domains

Let \(M\) be a function with domain \([-7,-4) \cup [-2,1] \cup [1,7)\).

Below is the graph of \(y = M(d)\).

A new function is created by shifting \(M\).

The function \(H(k)\) is defined as \(H(k) = M(k+3)\) with the induced domain.

We can see that \(M(-7) = 1\). This tells us that \(H\left (\answer {-10}\right ) = \answer {1}\).

There is an open dot on the graph of \(y= M(d)\) when \(d = -4\). There will be a corresponding open dot on the graph of \(z = H(k)\) when \(k = \answer {-7}\).

There is a horizontal line segment in the graph of \(y = M(d)\) from \((-2, -7)\) to \((1, -7)\). There will be a horizontal line segment in the graph of \(z = H(k)\) from \(\left (\answer {-5}, -7\right )\) to \(\left (\answer {-2}, -7\right )\).

There is a closed dot on the graph of \(y= M(d)\) when \(d = 1\). There will be a corresponding closed dot on the graph of \(z = H(k)\) when \(k = \answer {-2}\).

There is an open dot on the graph of \(y= M(d)\) when \(d = 7\). There will be a corresponding open dot on the graph of \(z = H(k)\) when \(k = \answer {4}\).