My Math Forum Need help with some questions

 Linear Algebra Linear Algebra Math Forum

 November 8th, 2017, 06:29 AM #1 Newbie   Joined: Nov 2017 From: Brazil Posts: 1 Thanks: 0 Need help with some questions Hi ppl. Please I need some help with these questions. Considering the parable with the function f(x) = -x² + 4x - 3. How do I prove that the point (2,1) belongs to the parable? And, also: Given the linear transformation T: R2 -> R2 : T1 (x,y) = ( cosθ * x - senθ * y , senθ * x + cosθ * y Given the angle θ = 150º, determine the rotation and the root positions of the parable after the linear transformation Thanks in advance in any help that could be provided.
November 8th, 2017, 03:36 PM   #2
Senior Member

Joined: Sep 2015
From: USA

Posts: 1,692
Thanks: 860

Quote:
 Originally Posted by Luke 2087 Hi ppl. Please I need some help with these questions. Considering the parable with the function f(x) = -x² + 4x - 3. How do I prove that the point (2,1) belongs to the parable? And, also: Given the linear transformation T: R2 -> R2 : T1 (x,y) = ( cosθ * x - senθ * y , senθ * x + cosθ * y Given the angle θ = 150º, determine the rotation and the root positions of the parable after the linear transformation Thanks in advance in any help that could be provided.
a point that lies on a curve given by a function $f(x)$ will be of the form

$(x, f(x))$

so let's take a look

$f(2) = -(2)^2 + 4(2) - 3 = -4 + 8 - 3 = 1$

so yes, the point $(2,1)$ lies on the parabola given by $f(x)$

for the 2nd bit first you have to find the roots. There will be 2 of them, call them $r_1,~r_2$

the points will then be $(r_1, 0), ~(r_2, 0)$

and just apply T to those 2 points.

 Tags questions

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post vickyc95 Number Theory 4 May 24th, 2016 08:20 PM Valar30 Calculus 17 January 4th, 2011 08:58 PM RMG46 Algebra 17 June 20th, 2010 10:31 PM tobymac Algebra 2 June 12th, 2009 10:33 AM blackobisk Algebra 1 January 27th, 2009 12:15 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top