7.4 Graphing Exponential, Logarithmic, and Cube Root Functions

7.4 Graphing Exponential, Logarithmic, and Cube Root Functions

A graphing item is scored on features, not artistry. Every sketch below is built from the same short checklist: anchor points, intercepts, asymptote, and end behavior.

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1. The Three Parent Graphs

When $0 < b < 1$, the exponential graph is decreasing and the logarithmic graph is decreasing as well, but every intercept and asymptote listed above is unchanged.

The Reflection Relationship

Because $y = \log_b(x)$ is the inverse of $y = b^x$, their graphs are reflections of each other across the line $y = x$. That single fact regenerates the whole logarithm row from the exponential row:

• The exponential's $y$-intercept $(0, 1)$ reflects to the logarithm's $x$-intercept $(1, 0)$.
• The exponential's horizontal asymptote $y = 0$ reflects to the logarithm's vertical asymptote $x = 0$.
• The exponential's range $(0, \infty)$ becomes the logarithm's domain $(0, \infty)$.

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2. Transformations and Where the Asymptote Goes

For $y = a\, b^{\,x-h} + k$ and $y = a \log_b(x - h) + k$ and $y = a\sqrt[3]{x - h} + k$:

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3. Worked Sketch 1: $y = 2^{\,x+1} - 4$

• Rewrite in transformation form. $x + 1 = x - (-1)$, so $h = -1$ and $k = -4$: shift left 1, down 4.
• Asymptote. $y = -4$. Draw it as a dashed line first; the curve approaches it from above as $x \to -\infty$ and never touches it.
• $y$-intercept. $y = 2^{0+1} - 4 = 2 - 4 = -2$, giving $(0, -2)$.
• $x$-intercept. Set $2^{\,x+1} - 4 = 0$, so $2^{\,x+1} = 4 = 2^2$, so $x + 1 = 2$ and $x = 1$, giving $(1, 0)$.
• A third anchor. $x = 2$ gives $y = 2^3 - 4 = 4$, so $(2, 4)$ - useful because it establishes the growth on the right.
• End behavior. As $x \to -\infty$, $y \to -4^+$. As $x \to \infty$, $y \to \infty$.

        y
      4 +                          * (2, 4)
        |                        /
      0 +----------------*------/-------------> x
        |          (1,0) |     /
     -2 +  ..........*(0,-2)
     -4 +- - - - - - - - - - - - - - - -   asymptote y = -4

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4. Worked Sketch 2: $y = \log_3(x - 2) + 1$

• Asymptote. The argument must be positive: $x - 2 > 0$, so the vertical asymptote is $x = 2$ and the domain is $(2, \infty)$.
• Anchor from the parent's $(1, 0)$. Set $x - 2 = 1$, so $x = 3$; then $y = 0 + 1 = 1$, giving $(3, 1)$.
• Anchor from the parent's $(b, 1)$. Set $x - 2 = 3$, so $x = 5$; then $y = 1 + 1 = 2$, giving $(5, 2)$.
• $x$-intercept. Set $\log_3(x - 2) + 1 = 0$, so $\log_3(x - 2) = -1$, so $x - 2 = 3^{-1} = \tfrac{1}{3}$ and $x = \tfrac{7}{3} \approx 2.33$.
• End behavior. As $x \to 2^+$, $y \to -\infty$. As $x \to \infty$, $y \to \infty$, but slowly: reaching $y = 3$ requires $x = 11$.

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5. Worked Sketch 3: $y = \sqrt[3]{x + 8} - 1$

• No asymptote and no domain restriction. Cube roots accept every real number, so the curve extends across the entire plane.
• Inflection point. The parent's inflection at the origin moves to $(h, k) = (-8, -1)$. This is the point of $180^\circ$ rotational symmetry for the transformed graph.
• $y$-intercept. $y = \sqrt[3]{0 + 8} - 1 = 2 - 1 = 1$, giving $(0, 1)$.
• $x$-intercept. Set $\sqrt[3]{x + 8} = 1$, so $x + 8 = 1$ and $x = -7$, giving $(-7, 0)$.
• Symmetric partner. The point $(0, 1)$ sits $8$ units right of and $2$ units above the inflection point $(-8, -1)$. Odd symmetry means the mirrored point sits $8$ units left of and $2$ units below it, at $(-16, -3)$. Check: $\sqrt[3]{-16 + 8} - 1 = \sqrt[3]{-8} - 1 = -2 - 1 = -3$.
• End behavior. As $x \to -\infty$, $y \to -\infty$; as $x \to \infty$, $y \to \infty$. The curve is always increasing and flattens away from the inflection point rather than approaching any line.

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6. The Grader's Checklist

Common deductions: drawing a horizontal asymptote at $y = 0$ on an exponential that was shifted vertically; letting a logarithmic curve cross its vertical asymptote; and drawing a cube root as if it stopped at its inflection point, which is a square root behavior, not a cube root one.