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These three terminologies are starting to confuse me.

To calculate the inverse, to my understanding, you must complete the square or use the quadratic equation to factor, then solve for y.

y = ax^2 + bx + c (quadratic)
y = a(x – h)2 + k (vertex form)
y = root(-k/a)-h (inverse)

To calculate the roots, to my understanding, you must complete the square or use the quadratic equation to factor, then solve for y,

x = (quadratic formula)

Similarly, with the y intercept, to my understanding, you must factor with the above methods and then solve for y.

I'm firstly curious why factoring is necessary at all; algebraically speaking, this step shouldn't be necessary.

I'm also curious what the relationship is between these three terms. If they are all solved the same way, are they all the same?

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  • $\begingroup$ Are you referring to the quadratic form $y = ax^2 + bx + c$? Could you give an example on each part? $\endgroup$
    – trang1618
    Commented Sep 27, 2016 at 19:43
  • $\begingroup$ examples coming $\endgroup$ Commented Sep 27, 2016 at 19:46
  • $\begingroup$ with the y intercept ... you must factor with the above methods and then solve for y The y-intercept is usually defined as the $(0,y)$ point of intersection with the y-axis. To find the $y$ of the intercept just set $x=0$ in the definition of the function (for a polynomial, this will give the free term). As for the other questions, the roots are numbers (actually the x-intercepts of the polynomial), while the inverse is a function. So, no, the three terms are not the same at all. $\endgroup$
    – dxiv
    Commented Sep 27, 2016 at 19:46

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I'm not sure what role $y$ plays here, so I'm guessing a bit. First, a quadratic would be written $ax^2+bx+c$. I can set this equal to $y$ or to $0$. If I set it equal to $0$, I get $0=ax^2+bx+c$, which I can solve by factoring or by quadratic formula or by completing the square. I'd get $x=r$ and $x=s$ for some numbers $r$ and $s$. The roots of the quadratic are $r$ and $s$. (They are also the $x$-intercepts of the curve $y=ax^2+bx+c$. We get $x$-intercepts by setting $y=0$. It could be that $r$ and $s$ are equal to each other or are complex numbers.)

If I set it equal to $y$, I get $y=ax^2+bx+c$. The $y$-intercept is the point where $x=0$, so just plug in $x=0$ and get $y=c$.

The "inverse" of this function won't exist unless you restrict the domain of $x$ so that the function is one-to-one. If you complete the square and get $y=a(x-g)^2+h$, restrict $x\geq g$ and solve for $x$ you'll have the inverse function. Something like:

$$y-h = a(x-g)^2 \implies (x-g)^2 = \frac{y-h}{a} \implies x-g = \sqrt{\frac{y-h}{a}} $$

where we take the positive square root because $x\geq g$. So $$x=g+\sqrt{\frac{y-h}{a}} $$ gives a formula for the inverse, although most people would then swap the roles of $x$ and $y$.

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  • $\begingroup$ I guess my point is that if you remove the x value in an inverse function, you get the formula for it's roots. $\endgroup$ Commented Sep 27, 2016 at 22:23
  • $\begingroup$ I'm going to make this another question. $\endgroup$ Commented Sep 27, 2016 at 22:24
  • $\begingroup$ math.stackexchange.com/questions/1944217/… $\endgroup$ Commented Sep 27, 2016 at 22:33

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