I have been asked to find the inverse of an equation that has the form
$y=ax^2 + by -c$

EDIT: Which is $y=4x^2+ 8x -3$ in the graph below

Using a graphing calculator, and trial and error, I can find the equation in the form of

$y=af(k(x-d)^2)+ c$

EDIT: Which is $y=4(x+1)^2-7$ in the graph below.

NOTE: This equation has been graphed, and falls directly below the $y=4x^2+8x-3$ (because they are the equal)

From this notation, I can then easily find the inverse.

enter image description here

However, I cannot seem to figure out how to do this algebraically. I am told I should 'complete the square'. I have used multiple inverse calculators on many websites and the all have given me different steps/answers.

How can I convert this equation into the form $y=af(k(x-d))+c$ algebraically?

(Seen above as $y=4(x+1)^2-7$ )


To convert from $4x^2+ 8x -3$ to $4(x+1)^2-7$, I believe I am just finding the vertex and then applying the transformations from 0,0.

  • 2
    $\begingroup$ $a x + b y - c$ defines a function of 2 variables, which you can't graph in the 2D plane. $a x + b y - c = 0$ defines a line in the plane. Neither interpretation has any square to complete so it's hard to understand what you are asking. $\endgroup$
    – dxiv
    Commented Sep 19, 2016 at 22:39
  • $\begingroup$ Sorry. Fixed the question. $\endgroup$ Commented Sep 19, 2016 at 22:59
  • $\begingroup$ The title and first line still refer to $a x + b y - c$. Also, an equation is not a function, and does not have an inverse function. If this is homework, please copy the question literally, and show your attempts at solving it. You may find Completing the Square useful. $\endgroup$
    – dxiv
    Commented Sep 19, 2016 at 23:08
  • $\begingroup$ Wouldn't the inverse of an expression poo (x,y) simply be poo (y,x)? Wouldn't this inverse be ay+bx-c. If not what is the definition of inverse? $\endgroup$
    – fleablood
    Commented Sep 19, 2016 at 23:27
  • $\begingroup$ Dvix ax+by -c isn't even an equation. $\endgroup$
    – fleablood
    Commented Sep 19, 2016 at 23:28

2 Answers 2


The suggestion you received is good, you shall proceed as to "absorb" the factor $x$: $$ y = 4x^{\,2} + 8x - 3 = 4x^{\,2} + 8x + 4 - 4 - 3 = 4\left( {x + 1} \right)^{\,2} - 7 $$ and then, clearly $$ \left( {x + 1} \right)^{\,2} = \frac{{y + 7}} {4}\quad \Rightarrow \quad x = \pm \frac{1} {2}\sqrt {y + 7} - 1 $$ Of course, this is not always viable in general.

  • $\begingroup$ I can't seem to get it to work when b=2 $\endgroup$ Commented Sep 20, 2016 at 2:24

If you mean the function $y=ax^2+bx-c$, then the inverse function will be: $$x=ay^2+by-c\\ay^2+by-c-x=0\\y_{1,2}=\frac{-b\pm\sqrt{b^2+4a(c+x)}}{2a}$$ So your inverse function is actually 2 functions.

  • $\begingroup$ While this is insightful, part of the question requires completing the square, not using the quadratic formula. $\endgroup$ Commented Sep 19, 2016 at 23:56
  • $\begingroup$ Then do it by completing the square. The idea is to solve it when it has been inverted. Nobody (except your teacher) cares how you solve it. So solve it by completing the square. $\endgroup$
    – fleablood
    Commented Sep 19, 2016 at 23:58
  • $\begingroup$ Basically, the whole formula, is completing the square. $\endgroup$
    – 76david76
    Commented Sep 20, 2016 at 4:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .