Wemazi

Topic 1.9

Choose a formula with the required behavior

Builds on locating a hole and its limiting value. 6 minutes.

Build a hole at (−2, 5)

You need a formula whose graph has an open ring at \((-2,5)\). The curve \(y=x^2+1\) passes through this point, which the orange dot marks.

Press the button to multiply the top and the bottom by \(x+2\). The formula becomes \(\dfrac{(x^2+1)(x+2)}{x+2}\).

At every input except \(-2\), the factor \(x+2\) cancels, so the graph is still \(y=x^2+1\). At \(x=-2\) the bottom is 0, so the formula has no value there, and the orange dot becomes an open ring.

A hole anywhere you choose

Drag the orange point to choose where the hole goes, at the input \(c\) and the height \(L\). The curve \(g\) passes through it with no break, so \(g(c)=L\).

Press the button to multiply the top and the bottom of \(g\) by \(x-c\).

The new formula equals \(g\) at every input except \(c\), so its values near \(c\) approach \(g(c)\), which is \(L\). At \(c\) the bottom is 0, so the formula has no value there, and the point becomes a ring.

Wherever the point is, the hole comes from a curve through the point and the factor x − c in the top and the bottom.

Notation

  1. Choose a formula g whose graph passes through the required point with no break.

    \[g(c)=\cy{L}\]

  2. Multiply the top and the bottom of \(g\) by \(x-c\).

    \[f(x)=\frac{g(x)\,\cq{(x-c)}}{\cq{x-c}}\]

  3. The factor cancels when \(x\neq c\), so \(f\) equals \(g\) there.

    \[f(x)=g(x)\quad (x\neq c)\]

  4. At c the bottom is 0, so f has no value there, and its graph has a ring at the required point.

    \[(c,\ \cy{L})\]

When it fails

The formula is now \(\dfrac{x^2+1}{x+2}\). At \(x=-2\) the top is 5, not 0, so nothing cancels, and the values grow without bound near \(-2\).

A formula is bounded near c when its values stay between two fixed numbers. With the factor only in the bottom, the formula is not bounded near c, and the graph has no hole.

Example 1: a limit and a separate value

Write a formula \(f\) with \(\displaystyle\lim_{x\to 1} f(x)=6\) and \(f(1)=2\).

  1. Choose a formula that equals 6 at \(x=1\). The line \(3x+3\) works.

    \[3(1)+3=\cy{6}\]

  2. Use the line for every x except 1. The two orange dots near 1 are close to height 6, so the limit is 6.

  3. Give the value at 1 as a second case of the formula. This puts a chalk dot at (1, 2), below the ring.

    \[f(x)=\begin{cases}3x+3 & x\neq 1\\ 2 & x=1\end{cases}\]

Example 2: a hole and no other break

Write a formula whose graph has a hole at \((1,-3)\) and no other break.

  1. Choose a curve through \((1,-3)\). The curve \(y=x^2-4\) works.

    \[1^2-4=\cy{-3}\]

  2. A tempting wrong move: Put \(x-1\) in the bottom only, to make the formula undefined at 1.

    \[f(x)=\frac{x^2-4}{x-1}\]

    The top is \(-3\) at \(x=1\), not 0, so nothing cancels, and at \(x=1.01\) the value is about \(-298\). The red graph has a dashed vertical line and no ring.

  3. Put \(x-1\) in the top as well, so that it cancels. The graph is \(y=x^2-4\) with one ring at \((1,-3)\) and no other break.

    \[f(x)=\frac{(x^2-4)\,\cq{(x-1)}}{\cq{x-1}}\]

Practice

Write a fraction of two polynomials whose graph is the line \(y=2x+4\) with a hole at \((3,10)\).

Move the slider to x = 3 and read the height of the line. Then multiply the top and the bottom by x − 3.

Solution

  1. The line is at height 10 at \(x=3\), so the hole is on the line.

    \[2(3)+4=\cy{10}\]

  2. Multiply the top and the bottom by \(x-3\). The graph is the same line with a ring at \((3,10)\).

    \[f(x)=\frac{(2x+4)\,\cq{(x-3)}}{\cq{x-3}}\]

  3. Expand the top to write one fraction of two polynomials.

    \[f(x)=\frac{2x^2-2x-12}{x-3}\]

Questions

The first graph is the line \(y=x+2\) with an open ring at \((1,3)\). The second graph is an orange curve with a dashed vertical line at \(x=1\).

  1. Which formula has the first graph?

    • \(\dfrac{x^2+x-2}{x-1}\)
    • \(\dfrac{x+2}{x-1}\)
    • \(\dfrac{x^2-1}{x-1}\)
    • \(x+2\)
    Answer

    \(\dfrac{x^2+x-2}{x-1}\)

  2. The second graph is \(\dfrac{x+2}{x-1}\). Which change gives the first graph?

    • Multiply the top by \(x-1\)
    • Add \(x-1\) to the top
    • Remove the bottom
    • Change the bottom to \(x+1\)
    Answer

    Multiply the top by \(x-1\)

  3. Which formula has the limit 7 as \(x\) approaches 3 and has no value at \(x=3\)?

    • \(\dfrac{x^2+x-12}{x-3}\)
    • \(\dfrac{x+4}{x-3}\)
    • \(\dfrac{x^2-9}{x-3}\)
    • \(\dfrac{x^2+x-12}{x+4}\)
    Answer

    \(\dfrac{x^2+x-12}{x-3}\)

Summary

To build a hole at \((c,L)\), you choose a curve through \((c,L)\) with no break and multiply its top and its bottom by \(x-c\).

This works because \(x-c\) cancels at every input except \(c\), so the graph keeps the curve and loses one point.

If a separate value at c is also required, you give it as a second case of the formula.

This is one of four skills in topic 1.9 of Which & Why: AP Calculus.