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.
Best Answer
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.