Skip to main content
Chemistry LibreTexts

5.9: Putting It Together- Nonlinear Models

  • Page ID
    251346
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

     

    Let’s Summarize

    This is what we have learned about exponential models:

    The general form of an exponential model is y = C · bx.

    • Exponential models are nonlinear. More specifically, exponential models predict that y increases or decreases by a constant percentage for each 1-unit increase in x.
    • C is the initial value. It is the y-value when x = 0. It is also the y-intercept.
    • b is the growth factor or decay factor. b is always positive.
      • If b is greater than 1, b is a growth factor. In this case, the association is positive, and y is increasing. This makes sense because multiplying by a number greater than 1 increases the initial value. From the growth factor, we can determine the percent increase in y for each additional 1-unit increase in x.
      • Similarly, if b is greater than 0 and less than 1, b is a decay factor. In this case, the association is negative, and y is decreasing. From the decay factor, we can determine the percentage decrease in y for each additional 1-unit increase in x.

    Let’s compare the general form of an exponential model to the general form for a linear model:y = a + bx.

    • In the linear model, a is the initial value. It is the y-value when x = 0. It is also the y-intercept.
    • b is the slope. From the slope, we can determine the amount and direction the y-value changes for each additional 1-unit increase in x. When b is positive, there is a positive association, and y increases. When b is negative, there is a negative association, and y decreases.
    CC licensed content, Shared previously

    5.9: Putting It Together- Nonlinear Models is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?