I'm teaching this year out of "Precalculus with limits" by Ron Larson [7th ed], and the following expression appears in the unit introducing polynomial functions:


He calls this the "standard form of a quadratic function".

I've only ever seen this form referred to as "vertex form", whereas "standard form" instead means a quadratic expression written as:


This ambiguity is EXTREMELY confusing for my students, and it's really obnoxious for me, too. (I have to rewrite all materials to

Does anyone know why this terminological variation is introduced in this textbook or how well-established it is in general? I can't think of any good reason for changing it now unless it actually is well-established elsewhere. But I don't think I've ever seen this over the past 25 years, at least not in the US.

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    $\begingroup$ FYI, on college algebra and precalculus tests I almost never used phrases such as "vertex form" or "standard form" or "slope-intercept form" (for lines). Instead, I would use $a(x-h)^2 + k$ or $ax^2 + bx + c$ or $y=mx+b$ (or different letters, if necessary to agree with the text). However, this rarely came up anyway, since my questions would be to find the vertex and axis of symmetry, or graph it (this being for non-graphing calculator tests). Usually only in the case of lines would the equation be an answer (and it would be "$y=mx + b$ or $x=c$ form," this also for tangent lines in calculus. $\endgroup$ Nov 11, 2022 at 17:17
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    $\begingroup$ Reminds me of the joke: "The nice thing about standards is that there are so many to choose from!". But really, "standard" isn't a very good descriptor as it doesn't tell you much about why you would do it or what properties it has. $\endgroup$
    – Adam
    Nov 12, 2022 at 15:06
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    $\begingroup$ I fear that the question is not well suited for this platform since no one except the author can explain why they introduced the terminology this way. $\endgroup$
    – Jasper
    Nov 13, 2022 at 13:34
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    $\begingroup$ Well, yes, you have a point. But it's possible that someone has read other works by this author or attended discussions with him or others where this issue has been addressed. The textbook is already up to its 7th edition by a major publisher, so presumably it has a significant following. Also, note that there is also a question of the more general establishment of this usage apart from this specific author. If this forum isn't appropriate for asking other educators to comment on the basis of their experience, then what is it for? $\endgroup$
    – Cassius12
    Nov 14, 2022 at 18:10
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    $\begingroup$ @SueVanHattum: Stewart's Precalculus indeed call $f(x)=a(x-h)^2+k$ standard form. $\endgroup$
    – Taladris
    Nov 15, 2022 at 8:27

1 Answer 1


When considering transformations of the "basic" $y=x^2$, there is not much else you can do besides

  • Vertical shift

  • Horizontal shift

  • Scaling

The first is achieved by adding a constant to $y$. The second is achieved by adding a constant to $x$. The third can be achieved in several ways, such as by a ratio of coefficients applied to $x$ and $y$ separately.

This leads to something in the form $m\cdot y+a={(n\cdot x+b)}^2$. It is an exercise in manipulation and renaming to obtain $y=k\cdot {(x-b)}^2+c$ from there.

The same transformations (and in a teaching sequence, for consistency between symbols and actions, the letters) can be applied to cubics and quartics, hyperbolae and sines, exponential and logarithm curves, absolutes and ceilings and floors...

One could and perhaps should consider these to be standard forms of the equations matching the transformed graphs involved. While it makes less sense if one is only working in polynomials or where transformations are a secondary concern, it is a coherent position to hold when working graphically and when using the equations for modelling (since it allows for direct correspondence between the coefficients/constants and aspects of many models).


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