Is it really necessary to learn how to write a quadratic in standard to vertex form?

Question

How crucial is this skill or form of writing and polynomial function? Can't we just always use the $-b/2a$ trick for the $x$ intercept and just plug it back into the function to find the $y$?

Answer 1

The way your question is written leaves me uncertain what your question is.

What is "standard to vertex form"?

One can write $y = ax^2 + bx + c$ and show that that is equivalent to $y= a\left( x + \dfrac b {2a} \right)^2 + c - \dfrac{b^2}{4a}.$ Is the latter what you mean by "standard to vertex form"?

What is the $\dfrac{-b}{2a}$ trick? The number $\dfrac{-b}{2a}$ is the $x$-coordinate of the vertex. Did you have in mind that knowing that that is the $x$-coordinate of the vertex? At any rate, this doesn't give us any $x$-intercepts (as suggested in your question) without doing a bit more than that. Your phrase "the $x$-intercept" suggests that there is only one of those, and that's not true.

If my guesses as to your meaning are correct, it still leaves me wondering what you're asking. You seem to suggest that one can use the knowledge that the $x$-coordinate of the vertex is $-b/(2a)$ to do something that could otherwise be done by knowing that $ax^2+bx+c = a\left( x + \dfrac b {2a} \right)^2 + c - \dfrac{b^2}{4a},$ and that using the aforentioned knowledge is a preferable way to do it. But what is it that you're trying to do that could be done by either of those methods?

And how would one know that $-b/(2a)$ is the $x$-coordinate of the vertex without showing it by using the fact that $ax^2+bx+c = a\left( x + \dfrac b {2a} \right)^2 + c - \dfrac{b^2}{4a}\text{?}$

Answer 2

To answer your question, it isn't crucial in any serious way. However, it is as handy as, say, "point-slope form" or "slope-intercept form" for lines, it gives you a way to immediately recognize certain bits of information. Additionally, as Daniel M. mentions, it covers completing the square which is more often than not, the desired way to solve for the roots etc. of a quadratic (it's also just a handy algebraic manipulation.

Perhaps to demonstrate some utility of completing the square, we'll derive the quadratic formula starting from standard form:

$$ax^2+bx+c=0 \implies x^2+\frac{b}{a}x=-\frac{c}{a}$$

but now, we really need to complete the square:

$$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}-\frac{c}{a} \implies(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2} \implies x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

Notice how one could deduce the $\frac{-b}{2a}$ trick from the middle implication! You're right to notice that the two are intimately related, I think it's best to know both.

Let's use a similar technique to examine what's going on.

vertex form: $y=a_1(x-h)^2+k$. But, given a quadratic in standard form, $y=ax^2+bx+c$, we can deduce the following by adding and subtracting the same thing, while factoring out an $a$, and then completing the square:

\begin{align*}y&=ax^2+bx+c\\ &=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)+\left(\frac{b^2}{4a}-\frac{b^2}{4a}\right)\\ &=a\left(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}+\frac{c}{a}-\frac{b^2}{4a^2}\right)\\ &=a\left((x+\frac{b}{2a})^2+\frac{c}{a}-\frac{b^2}{4a^2} \right)\\ &=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}. \end{align*}

Notice that this is a way to always get to "vertex form" and tells you exactly what $a,h,k$ are supposed to be in vertex form.

Namely: $a_1=a$ (Hoorah!)

$h=-\frac{b}{2a}$ (As you have noticed)

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

Answer 3

The ability to go from standard form to vertex form is critical! When you want to solve a quadratic equation, the first technique you learn is factoring. But not every quadratic equation is factorable, so how do we solve those?

The vertex form of the equation is always solvable: $$ a(x-h)^2+k=0 \iff (x-h)^2=-\frac{k}{a} \iff x-h=\pm\sqrt{-\frac{k}{a}} \iff x=h\pm\sqrt{-\frac{k}{a}}. $$

So the ability to go from standard form to vertex form is what allows us to solve any quadratic equation.

Answer 4

It's important for the simple reason that, given

$ax^2 + bx + c$, it isn't immediately obvious what the graph looks like. Converting to vertex form, however, it is: $$a\left(x+\frac{b}{2a}\right)^2 + c - \frac{b^2}{4a}$$

You have your standard $x^2$ parabola scaled and shifted.

Why, then, don't we always have vertex form? This is because it is more natural and conventional to write polynomials as

$$P(x) = \sum_{i=0}^n a_ix^i$$

With this, for example, differentiation and integration is simpler. Plus vertex form, to my knowledge, does not generalize in an obvious way to polynomials of arbitrary degree.