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I was watching this video but at 8:05 I don't get why to solve for the function $p(t) = at^2 + bt + c \geq 0$, Sal decides to input $t= \frac{b}{2a}$. Someone made this explanation:

$\frac{b}{2a}$ is something that comes up a lot when solving quadratic equations. All quadratic equations form a parabola which is given by the equation: $$ax² + bx +c = y$$ All parabolas have a single vertex. The left and right sides of the equation are symmetric about the vertex, so if we can find the value at the vertex, we can easily find other information about the parabola and easily solve for $x$ given any value for $y$ (and $a$, $b$, and $c$ obviously). If we change our equation into the form: $$ax²+bx = y-c$$ Then we can factor out an $x$: $$x(ax+b) = y-c$$ Since $y-c$ only shifts the parabola up or down, it's unimportant for finding the $x$-value of the vertex. Because of this, I'll simply replace it with $0$: $$x(ax+b) = 0$$ Now, we just solve for $x$: $x = 0$; and $ax+b = 0\implies x = -b/a$. This gives us two values of $x$ that are an equal distance away from the vertex point. So, the vertex point is the value perfectly in between them (or the average). This gives: $$v_x = (0+(-b/a))/2$$ or $$v_x = -b/2a$$ ($v_x$ is the $x$-value of the vertex). If you have any function, you can shift it left or right by changing the input: $f(x-h)$ shifts the graph $f(x)$ to the right by $h$ units. So, when Sal inputs $\frac{b}{2a}$ into the equation, what he's doing is inputting the value that will shift the vertex point to $x=0$. It's somewhat complex, but hopefully this helps.

But would not then be $\frac{-b}{2a}$ so that is the vertex instead of $\frac{b}{2a}$?

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  • $\begingroup$ In the video the quadratic function obtained is $p(t)=at^2-bt+c$ (minus $b$!), where $a=y.y,b=2x.y,c=x.x$ that´s why he inserts $\frac{b}{2a}$, because the vertex has $\frac{b}{2a}$ as its $x$- coordinate, otherwise ($...+bt...$) You would have been right $\endgroup$ – Peter Melech Feb 9 '16 at 15:59
  • $\begingroup$ Thanks, I was so tired that did not see that, thats why I asked for help $\endgroup$ – kprincipe Feb 9 '16 at 17:58
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Note that

$$at^2+bt+c = a\bigg(t+\frac{b}{2a}\bigg)^2+c-\bigg(\frac{b^2}{4a}\bigg)$$

So $p$ attains a minimum of $\bigg(c-\frac{b^2}{4a}\bigg)$ at $t = -\frac{b}{2a}$.

To check this,

$$p\bigg(-\frac{b}{2a}\bigg) = a\frac{b^2}{4a^2}-b\frac{b}{2a}+c$$

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

So in general $t = -\frac{b}{2a}$ gives you the value of the independent variable where the function value is at a minimum or maximum. There is no 'shift' taking place.

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