For this quadratic function, I need to translate the $(x,y)$ coordinate axes horizontally and vertically, so that the new origin of the coordinate system is located at the point with old coordinates $(2,3)$.
The new $(X,Y)$ coordinates are therefore related to the old $(x,y)$ -system by the equations:
Question -1
Since the new coordinates is 2 and 3 for x and y respectively, I substract them from x and y. Even though it is correct when I enter into the system, I don't fully understand why I need to substract, why not addition.
If you could guide me for further reading or explanation, that would be great.
$X = x -2 $
$Y = y -3 $
$ y = 2x^2−3x+1$
I need to substitute these $X = x -2, Y = y -3$, into the formula $y = 2x^2−3x+1$, in order to express Y as a function of X.
Below is the work. $y = ax^2 + bx + c = a(x + b/2a)^2 + (c-b^2/4a)$
$y = 2x^2−3x+1
\
= 2( x - 3/4)^2 + (1-9/8)
\
= 2( x - 3/4)^2 -1/8
\
y + 1/8 = 2( x - 3/4)^2$
Let $Y= y + 1/8$
Let $X= x - 3/4$
Therefore $Y=2X^2$
Question -2
My calculation for Y is as below. However, when I enter this answer, I still get incorrect response. I have checked it many times, but still couldn't spot the mistake. Is my method incorrect?
$y -3 = 2(x-2)^2 - 3(x-2)+ 1$
$y = 2(x^2 -4x + 4) - 3x -6 + 1 + 3$
$y = 2x^2 -8x + 8 -3x -2$
$y = 2x^2 -11x + 6$
Question 1:
I do not know if i can satisfy your first question. Simply put the new center coordinates in your equation
$$ X= x-2 = 2-2 = 0, $$ $$ Y= y-3 = 3-3 = 0. $$
You see that $X$ an $Y$ are zero at the new center $x=2$ and $y=3$.
Or vice versa put the old center coordinates in the new system
$$ -2= x-2, $$ $$ -3= y-3, $$
resulting in
$$ x=-2+2=0, $$ $$ y=-3+3=0. $$
You see that $x$ an $y$ are zero at old center for $X=-2$ and $Y=-3$.
Question 2:
In the second line there is a typo
$$ y= 2 (x^2 -4 x + 4) - 3x + 6 + 1 + 3 .$$