How to go from general form to standard form of quadratic equation? (with a deeper question about forms as well)

Question

I'm learning how to convert quadratic equations from general form to standard form, in order to make them easier to graph. We know the general form is ax^2+bx^2+c, and the standard form is a(x-h)^2+k. To help with the conversion, we can expand the standard form, and see that it turns into the general form. I totally get how to go from standard to general. I can easily memorize what h and k are, and use them to consistently derive standard forms.

What I'm curious about is how to, a priori, go from the general form to the standard form? Is there a way to see that ax^2+bx+c can turn into a(x-h)^2+k without knowing that form ahead of time? How was the form a(x-h)^2+k discovered in the first place? How are alternate forms of equations discovered in general? I honestly wouldn't know where to begin.

I ask out of curiosity, and because I believe knowing how to go in the other direction will help really solidify this concept for me. Even if that knowledge is above my skillset at the moment, at least an overview of what kind of math is involved may supplement this concept for me.

Answer 1

To get the standard form from the general form, first factor out $a$, then complete the square, and finally adjust the constant term:

$$\begin{align*} ax^2+bx+c&=a\left(x^2+\frac{b}ax+\frac{c}a\right)\\ &=a\left(\left(x+\frac{b}{2a}\right)^2+u\right)\,,\tag{1} \end{align*}$$

where I don’t yet know what $u$ is. To find out, I multiply out $(1)$ to get

$$a\left(x^2+\frac{b}ax+\frac{b^2}{4a^2}\right)+au\,,$$

which is

$$ax^2+bx+\frac{b^2}{4a}+au\,.$$

If this is to be the same as the original quadratic, we must have

$$\frac{b^2}{4a}+au=c\,,$$

meaning that

$$u=\frac{c-\frac{b^2}{4a}}a=\frac{c}a-\frac{b^2}{4a^2}\,,$$

and our quadratic can be written as

$$a\left(x+\frac{b}{2a}\right)^2+\left(c-\frac{b^2}{4a}\right)\,.$$

Thus, the $h$ and $k$ of the standard form must be $$h=-\frac{b}{2a}$$ and $$k=c-\frac{b^2}{4a}\,.$$

Take the quadratic $2x^2+3x+4$ as an example. Factoring out the leading coefficient of $2$ results in

$$2\left(x^2+\frac32x+2\right)\,.$$

Then

$$x^2+\frac32x+2=\left(x+\frac34\right)^2+u$$

for some $u$, and since $\left(x+\frac34\right)^2=x^2+\frac32x+\frac{9}{16}$, $u$ must be $2-\frac9{16}=\frac{23}{16}$. Thus,

$$\begin{align*} 2x^2+3x+4&=2\left(\left(x+\frac34\right)^2+\frac{23}{16}\right)\\ &=2\left(x-\left(-\frac34\right)\right)^2+\frac{23}4\,, \end{align*}$$

with $h=-\frac34$ and $k=\frac{23}8$.

Answer 2

Brian M. Scott's answer gives a good algebraic explanation; let me try to give a geometric summary of what's going on.

Remember that there are four basic "geometric transformations" of the graph of an equation $y=f(x)$. Specifically, for a constant $u$, we have:

• The graph of $y=f(x-u)$ is the graph of $y=f(x)$ shifted to the right by $u$.

• The graph of $y=f(ux)$ is the graph of $y=f(x)$ "compressed horizontally" by a factor of $u$ (and possibly reflected about the $y$-axis, depending on the sign of $u$).

• The graph of $y=f(x)+u$ is the graph of $y=f(x)$ shifted up by $u$.

• The graph of $y=uf(x)$ is the graph of $y=f(x)$ "stretched vertically" by a factor of $u$ (and possibly reflected about the $x$-axis, depending on the sign of $u$).

Now - granting that a parabola is a transformation of the graph of $y=x^2$ via stretching, shifting, and flipping - this tells us that every parabola has the form $$y=u_0(u_1x-u_2)^2+u_3$$ for some $u_0,u_1,u_2,u_3$. Moreover, at least for parabolas (this doesn't work in general) we can combine two of these variables into a single one: the expression above is equivalent to $$y=(u_0u_1^2)(x-{u_2\over u_1})^2+u_3.$$ So we really only need three "transformation constants:"

Every parabola is the graph of an equation of the form $$y=w_0(x-w_1)^2+w_2$$ for some constants $w_0,w_1,w_2$.

And at this point there's just one last step remaining before the transformation into standard form is something guessable: realizing that the graph of $y=ax^2+bx+c$ is a parabola. This is something you can notice just by graphing a few examples (always graph a few examples!).