# Find inverse of a quadratic polynomial by 'completing the square'

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.

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

EDIT:

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.

• $a x + b y - c$ 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.
– dxiv
Commented Sep 19, 2016 at 22:39
• Sorry. Fixed the question. Commented Sep 19, 2016 at 22:59
• 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.
– dxiv
Commented Sep 19, 2016 at 23:08
• 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? Commented Sep 19, 2016 at 23:27
• Dvix ax+by -c isn't even an equation. Commented Sep 19, 2016 at 23:28

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.
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.