Translation

Algebra Level 3

You'd like to translate a parabola y = a x 2 + b x + c y=ax^2+bx+c horizontally, for a vector t = ( t , 0 ) , t R \vec{t}=(t,0),\: t\in \mathbb{R} . \newline New parabola is y t = a 2 x 2 + b 2 x + c 2 . y_t=a_2x^2+b_2x+c_2.\newline What is c 2 c_2 in terms of a , b , c a,b,c ?

c b t + a t 2 2 c-bt+\frac{at^2}{2} c + a t 2 2 c+\frac{at^2}{2} c + ( a b ) t c+(a-b)t c b t + a t 2 c-bt+at^2 c b t c-bt

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1 solution

Lovro Cupic
Aug 22, 2019

We'll take a 2 = a a_2=a for granted. The symmetry axis x = b 2 a x=\dfrac{-b}{2a} will translate to x t = b 2 a + t = b 2 2 a 2 x_t=\dfrac{-b}{2a}+t=\dfrac{-b_2}{2a_2} so we obtain b 2 = b 2 a t b_2=b-2at . Discriminant won't change because distance between solutions don't change with horizontal translation (equivalent condition would emerge by taking the minimal or maximal value as invariant under horizontal translation). Either way, b 2 4 a c = b 2 2 4 a c 2 b^2-4ac=b_2^2-4ac_2 . Inserting b 2 = b 2 a t b_2=b-2at yields c 2 = c b t + a t 2 c_2=c-bt+at^2 .

Better alternative is to notice that c 2 = y ( t ) c_2=y(-t) .

lovro cupic - 1 year, 9 months ago

Translating a curve in a fixed reference system is equivalent to translating the reference system in the opposite direction keeping the curve fixed in position . In the transformed reference system, the coordinates of a point are : x t = x + t x_t=x+t and y t = y y_t=y . So the equation of the curve y = f ( x ) y=f(x) in the original system should be y = f ( x + t ) y=f(x+t) in the transformed system. Hence c 2 c_2 in this case should be a t 2 + b t + c at^2+bt+c , isn't it?

A Former Brilliant Member - 1 year, 9 months ago

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Yes, that's why added this solution as "a better alternative" in the comment.

lovro cupic - 1 year, 8 months ago

substituting x t x-t for x x performs the translation: y t = a ( x t ) 2 + b ( x t ) + c = a x 2 + ( b 2 a t ) x + a t 2 b t + c y_t=a(x-t)^2+b(x-t)+c=ax^2+(b-2at)x+at^2-bt+c

so we get a 2 = a a_2=a , b 2 = b 2 a t b_2= b-2at , and c 2 = a t 2 b t + c \boxed{c_2=at^2-bt+c} .

K T - 1 year, 4 months ago

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