Questions   
Sort: 
 #1
avatar+107 
0
Feb 15, 2018
 #1
avatar+484 
+1
Feb 15, 2018
 #2
avatar+118690 
+3

Hi all,

Supermanaccz has asked me to come back an talk about this ancient question.

Thanks for your interest Supermanaccz    laugh

 

Firstly the question is not displaying properly because the coding used on this site has changed a lot in the past 3 and a bit years.

So I will try to put the question up how it may originally have been. I believe a diagram like the one below would have been included in the original question.  Thie is the graph of   y=f(x).

I have drawn it in a program called GeoGebra. GeoGebra is a great tool. I found it a little hard to use at first (especially for more complicated things) but if you persist with it it will become much easier. Geogebra is a free download. I use it often.

 

Now to the actual question:

 

The graph of  is shown below. Assume the domain of  is  [-4,4] and that the vertical spacing of grid lines is the same as the horizontal spacing of grid lines. 

Part (a): The points  (a, 4) and  (b, -4)  are on the graph of  y=f(2x)    Find  a  and  b

From the graph you can see that 2 points on the f(x)  graph are (-4,4) and (4,-4)

-------------------------------------

Here is what i said before:

Part a       y=f(2a)

f(-4)=4   but    f(2a)=-4    so   2a=-4     so   a=-2

f(4)=-4   but    f(2b)=4     so    2b=4      so    b=2

----------------------------------------

But in the next post I will do an entirley different example, perhaps you will be able to see how many graphs are related to parent graphs. In this case f(x) is the parent function and f(2x) it the 'offspring'  Multiplying the x by 2 will cause the graph to have half the width (from the y axis). 

Part (b): Find the graph of  y=2x         Verify that your points from part (a) are on the graph. 

Part (c): The points (c,4)  and (d,-4)  are on the graph of y=f(2x-8)     Find  c   and  d

Part (d): Find the graph of  y=f(2x-8)   Be sure to verify that your points from part (c) are on the graph both algebraically and geometrically. 

 

Feb 15, 2018
 #1
avatar+26396 
+1

What is the smallest positive integer that will satisfy the following congruences:

N mod 105 = 104,

N mod 111 = 110,

N mod 121 = 111,

N mod 122 = 111

Thanks for any help.

 

\(\begin{array}{|l|cll|} \hline \begin{array}{|rcll|} \hline \mathbf{ n } & \mathbf{\equiv} & \mathbf{104 \pmod{105}} \\ \mathbf{ n } & \mathbf{\equiv} & \mathbf{110 \pmod{111}} \\ \hline \end{array} \\ \begin{array}{lrcll} \Rightarrow & n -104 &=& 105x \\ & n -110 &=& 111y \\\\ \Rightarrow & n &=& 104+105x \\ & n &=& 110+111y \\\\ & n=104+105x&=& 110+111y \\ & 104+105x&=& 110+111y \\ & 105x-111y&=& 6 \quad & | \quad : 3 \\ & \mathbf{35x-37y}& \mathbf{=}& \mathbf{2} \qquad x,y \in Z \\ &\rightarrow \quad x &=& -1+37b^{^1)} \qquad b \in Z \\\\ & n &=& 104+105x \\ & n &=& 104+105(-1+37b) \\ & n &=& 104-105+105\cdot 37b \\ & n &=& -1+ 3885b \\ & \mathbf{n} &\mathbf{\equiv}& -1 \mathbf{\pmod{3885}} \qquad (1)\\ \end{array}\\ \hline \end{array}\)

\(\begin{array}{|l|cll|} \hline \begin{array}{|rcll|} \hline \mathbf{ n } & \mathbf{\equiv} & \mathbf{111 \pmod{121}} \\ \mathbf{ n } & \mathbf{\equiv} & \mathbf{111 \pmod{122}} \\ \hline \end{array} \\ \begin{array}{lrcll} \Rightarrow & n -111 &=& 121x \\ & n -111 &=& 122y \\\\ \Rightarrow & n &=& 111+121x \\ & n &=& 111+122y \\\\ & n=111x+121y&=& 111+122y \\ & 111x+121y&=& 111+122y \\\\ & \mathbf{121x-122y}&\mathbf{=}& \mathbf{0} \qquad x,y \in Z \\ &\rightarrow \quad x &=& 122a^{^2)} \qquad a \in Z \\\\ & n &=& 111+121x \\ & n &=& 111+121(122a) \\ & n &=& 111+121\cdot 122a \\ & n &=& 111+14762a \\ & \mathbf{n} &\mathbf{\equiv}& 111 \mathbf{\pmod{14762}} \qquad (2)\\ \end{array} \\ \hline \end{array} \)

 

After reducing, we have two formulas:

\( \begin{array}{|rcll|} \hline \mathbf{n} &\mathbf{\equiv}& -1 \mathbf{\pmod{3885}} \qquad &(1)\\ \mathbf{n} &\mathbf{\equiv}& 111 \mathbf{\pmod{14762}} \qquad &(2)\\ \hline \end{array} \)

 

\( \begin{array}{|rcll|} \hline \mathbf{n} &\mathbf{\equiv}& -1 \mathbf{\pmod{3885}} \\ \mathbf{n} &\mathbf{\equiv}& 111 \mathbf{\pmod{14762}} \\ \hline \end{array} \\ \begin{array}{lrcll} \Rightarrow & n +1 &=& 3885x \\ & n -111 &=& 14762y \\\\ \Rightarrow & n &=& -1+3885x \\ & n &=& 111+14762y \\\\ & n=-1+3885x&=& 111+14762y \\ & -1+3885x&=& 111+14762y \\ & \mathbf{3885x-14762y}& \mathbf{=}& \mathbf{112} \qquad x,y \in Z \\ &\rightarrow \quad x &=&3378+14762g^{^3)} \qquad g \in Z \\\\ & n &=& -1+3885x \\ & n &=& -1+3885\cdot (3378+14762g) \\ & n &=& -1+3885\cdot 3378+ 3885\cdot 14762g \\ & \mathbf{n} &\mathbf{=}& \mathbf{13123529 + 57350370g \qquad g \in Z } \\ \end{array}\)

 

The smallest positive integer is  \(\mathbf{13\ 123\ 529}\)

 

Proof:

\(\begin{array}{|rcll|} \hline \mathbf{13\ 123\ 529} &\text{mod } 105 =& 104\\ \mathbf{13\ 123\ 529} &\text{mod } 111 =& 110\\ \mathbf{13\ 123\ 529} &\text{mod } 121 =& 111 \\ \mathbf{13\ 123\ 529} &\text{mod } 122 =& 111 \\ \hline \end{array} \)

 

\(\begin{array}{lcll} \hline ^1) \\ \text{Solve of the diophantine equation $35x - 37y = 2$ } \\ \text{The variable with the smallest coefficient is $x$. The equation is transformed after $x$: }\\ \begin{array}{rcll} 35x &=& 2 + 37y \\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 2 + 37y } {35}} \\ &=& \dfrac{ 2 +35y+2y } {35} \\ &=& \dfrac{ 35y + 2 +2y } {35} \\ &=& \dfrac{35y}{35} + \dfrac{ 2 +2y } {35} \\ x &=& y + \dfrac{ 2 +2y} {35} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &a &=& \dfrac{ 2 + 2y } {35} \\ & 35a &=& 2 + 2y \\ \end{array} \\ \text{The variable with the smallest coefficient is $y$. The equation is transformed after $y$: }\\ \begin{array}{rcll} 2y &=& -2 + 35a \\ \mathbf{y} &\mathbf{=}& \mathbf{\dfrac{ -2 + 35a } {2}} \\ &=& \dfrac{ -2 + 34a+a } {2} \\ &=& \dfrac{ 34a - 2 +a } {2} \\ &=& -\dfrac{-2}{2} +\dfrac{34a}{2} + \dfrac{ a } {2} \\ y &=& -1+17a+ \dfrac{ a } {2} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &b &=& \dfrac{a} {2} \\ & 2b &=& a \\ \end{array} \\ \text{The variable with the smallest coefficient is $a$. The equation is transformed after $a$: }\\ \begin{array}{lrcll} \text{no fraction there:} & \mathbf{a} &\mathbf{=}& \mathbf{ 2b } \\ \end{array} \end{array} \)

 

\(\text{Elemination of the unknowns:}\\ \begin{array}{rcll} \mathbf{y} &\mathbf{=}& \mathbf{\dfrac{ -2 + 35a } {2}} \quad & | \quad \mathbf{a = 2b }\\ & = & \dfrac{-2 + 35\cdot (2b) } {2} \\ \mathbf{y} & \mathbf{=} & \mathbf{-1+35b} \\\\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 2 + 37y } {35}} \quad & | \quad \mathbf{y = -1+35b }\\ & = & \dfrac{ 2 + 37\cdot (-1+35b ) } {11} \\ \mathbf{x} & \mathbf{=} & \mathbf{-1 + 37b } \\ \end{array} \)

 

\(\begin{array}{lcll} \hline ^2) \\ \text{Solve of the diophantine equation $121x - 122y = 0 $ } \\ \text{The variable with the smallest coefficient is $x$. The equation is transformed after $x$: }\\ \begin{array}{rcll} 121x &=& 122y \\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 122y } {121}} \\ &=& \dfrac{ 121y+y } {121} \\ &=& \dfrac{121y}{121} + \dfrac{y} {121} \\ x &=& y + \dfrac{ y} {121} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &a &=& \dfrac{y} {121} \\ & 121a &=& y \\ \end{array} \\ \text{The variable with the smallest coefficient is $y$. The equation is transformed after $y$: }\\ \begin{array}{lrcll} \text{no fraction there:} & \mathbf{y} &\mathbf{=}& \mathbf{ 121a } \\ \end{array} \end{array} \)

 

\(\text{Elemination of the unknowns:}\\ \begin{array}{rcll} \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 122y } {121}} \quad & | \quad \mathbf{y=121a }\\ & = & \dfrac{122\cdot (121a) } {121} \\ \mathbf{x} & \mathbf{=} & \mathbf{122a} \\ \end{array}\)

 

\( \begin{array}{lcll} \hline ^3) \\ \text{Solve of the diophantine equation $3885x - 14762y = 112 $ } \\ \text{The variable with the smallest coefficient is $x$. The equation is transformed after $x$: }\\ \begin{array}{rcll} 3885x &=& 112 + 14762y \\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 112 + 14762y} {3885}} \\ &=& \dfrac{ 112 + 11655y + 3107y } {3885} \\ &=& \dfrac{11655y + 112 + 3107y }{3885} \\ &=& \dfrac{11655y }{3885} + \dfrac{112 + 3107y} {3885} \\ x &=& 3y + \dfrac{ 112 + 3107y } {3885} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &a &=& \dfrac{112 + 3107y } {3885} \\ & 3885a &=& 112 + 3107y \\ \end{array} \\ \text{The variable with the smallest coefficient is $y$. The equation is transformed after $y$: }\\ \begin{array}{rcll} 3107y &=& -112 + 3885a \\ \mathbf{y} &\mathbf{=}& \mathbf{\dfrac{ -112 + 3885a} {3107}} \\ &=& \dfrac{ -112 + 3107a + 778a } {3107} \\ &=& \dfrac{3107a - 112 + 778a }{3107} \\ &=& \dfrac{3107a }{3107} + \dfrac{- 112 + 778a} {3107} \\ y &=& 3a + \dfrac{ 112 + 3107a } {3107} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &b &=& \dfrac{-112 + 778a } {3107} \\ & 3107b &=& -112 + 778a \\ \end{array} \\ \text{The variable with the smallest coefficient is $a$. The equation is transformed after $a$: }\\ \begin{array}{rcll} 778a &=& 112 + 3107b \\ \mathbf{a} &\mathbf{=}& \mathbf{\dfrac{ 112 + 3107b} {778}} \\ &=& \dfrac{ 112 + 2334b + 773b } {778} \\ &=& \dfrac{2334b +112 + 773b }{778} \\ &=& \dfrac{2334b }{3107} + \dfrac{112 + 773b} {778} \\ a &=& 3b + \dfrac{ 112 + 773b} {778} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &c &=& \dfrac{112 + 773b } {778} \\ & 778c &=& 112 + 773b \\ \end{array} \\ \text{The variable with the smallest coefficient is $b$. The equation is transformed after $b$: }\\ \begin{array}{rcll} 773b &=& -112 + 778c\\ \mathbf{b} &\mathbf{=}& \mathbf{\dfrac{ -112 + 778c} {773}} \\ &=& \dfrac{ -112 + 773c + 5c } {773} \\ &=& \dfrac{773c -112 + 5c }{773} \\ &=& \dfrac{773c }{773} + \dfrac{-112 + 5c } {773} \\ b &=& c + \dfrac{ -112 + 5c} {773} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &d &=& \dfrac{-112 + 5c } {773} \\ & 773d &=& -112 + 5c \\ \end{array} \\ \text{The variable with the smallest coefficient is $c$. The equation is transformed after $c$: }\\ \begin{array}{rcll} 5c &=& 112 + 773d\\ \mathbf{c} &\mathbf{=}& \mathbf{\dfrac{ 112 + 773d } {5}} \\ &=& \dfrac{ 110+2 + 770d + 3d } {5} \\ &=& \dfrac{110+770d+2 + 5d }{5} \\ &=& \dfrac{110 }{5}+\dfrac{770d }{5} + \dfrac{2 + 3d } {5} \\ c &=& 22 + 154d + \dfrac{2 + 3d } {5} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &e &=& \dfrac{ 2 + 3d } {5} \\ & 5e &=& 2 + 3d \\ \end{array} \\ \text{The variable with the smallest coefficient is $d$. The equation is transformed after $d$: }\\ \begin{array}{rcll} 3d &=& -2 + 5e\\ \mathbf{d} &\mathbf{=}& \mathbf{\dfrac{ -2 + 5e } {3}} \\ &=& \dfrac{ -2 + 3e + 2e } {3} \\ &=& \dfrac{3e -2 + 2e}{3} \\ &=& \dfrac{3e }{3} + \dfrac{-2 + 2e} {3} \\ d &=& e + \dfrac{-2 + 2e } {3} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &f &=& \dfrac{ -2 + 2e } {3} \\ & 3f &=& -2 + 2e \\ \end{array} \\ \text{The variable with the smallest coefficient is $e$. The equation is transformed after $e$: }\\ \begin{array}{rcll} 2e &=& 2 + 3f\\ \mathbf{e} &\mathbf{=}& \mathbf{\dfrac{2 + 3f } {2}} \\ &=& \dfrac{ 2+2f+f } {2} \\ &=& \dfrac{2 }{2} + \dfrac{2f }{2} + \dfrac{f} {2} \\ e &=& 1 + f + \dfrac{f} {2} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &g &=& \dfrac{ f} {2} \\ & 2g &=& f \\ \end{array} \\ \text{The variable with the smallest coefficient is $f$. The equation is transformed after $f$: }\\ \begin{array}{lrcll} \text{no fraction there:} & \mathbf{f} &\mathbf{=}& \mathbf{ 2g } \\ \end{array} \end{array}\)

 

\(\text{Elemination of the unknowns:}\\ \begin{array}{rcll} \mathbf{e} &\mathbf{=}& \mathbf{\dfrac{ 2 + 3f } {2}} \quad & | \quad \mathbf{f=2g }\\ & = & \dfrac{2 + 3\cdot 2g } {2} \\ \mathbf{e} & \mathbf{=} & \mathbf{1 + 3g } \\\\ \mathbf{d} &\mathbf{=}& \mathbf{\dfrac{ -2 + 5e } {3}} \quad & | \quad \mathbf{e=1 + 3g }\\ & = & \dfrac{-2 + 5\cdot (1 + 3g) } {2} \\ \mathbf{d} & \mathbf{=} & \mathbf{1 + 5g } \\\\ \mathbf{c} &\mathbf{=}& \mathbf{\dfrac{ 112 + 773d } {5}} \quad & | \quad \mathbf{d=1 + 5g }\\ & = & \dfrac{112 + 773\cdot (1 + 5g) } {2} \\ \mathbf{c} & \mathbf{=} & \mathbf{177 + 773g } \\\\ \mathbf{b} &\mathbf{=}& \mathbf{\dfrac{ -112 + 778c } {773}} \quad & | \quad \mathbf{c=177 + 773g }\\ & = & \dfrac{-112 + 778\cdot (177 + 773g) } {773} \\ \mathbf{b} & \mathbf{=} & \mathbf{ 178 + 778g } \\\\ \mathbf{a} &\mathbf{=}& \mathbf{\dfrac{ 112 + 3107b } {778}} \quad & | \quad \mathbf{b=178 + 778g }\\ & = & \dfrac{112 + 3107\cdot (178 + 778g) } {778} \\ \mathbf{a} & \mathbf{=} & \mathbf{ 711 + 3107g } \\\\ \mathbf{y} &\mathbf{=}& \mathbf{\dfrac{ -112 + 3885a } {3107}} \quad & | \quad \mathbf{a=711 + 3107g }\\ & = & \dfrac{-112 + 3885\cdot (711 + 3107g) } {3107} \\ \mathbf{y} & \mathbf{=} & \mathbf{ 889 + 3885g } \\\\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 112 + 14762y } {3885}} \quad & | \quad \mathbf{y=889 + 3885g }\\ & = & \dfrac{112 + 14762\cdot (889 + 3885g) } {3885} \\ \mathbf{x} & \mathbf{=} & \mathbf{ 3378 + 14762g } \\ \end{array}\)

 

 

laugh

Feb 15, 2018
 #1
avatar+37157 
0
Feb 15, 2018

1 Online Users

avatar