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Given are the functions f(x)=-0,5x3+4x2-6x and g(x)=2-√(4+2x)

a. For what values of x does the function g have results?

b. Give the coordinates of the vertex of the graph of f

c. Plot the graphs of f and g and give the solution to g(x)≤f(x)

The question says that x and y are real numbers, it does not say that they are integers (necessarily).

The minimum is 3 + 6sqrt(2), approximately 11.585, at x = 1 + 3sqrt(2), y = 1 + 3sqrt(2)/2, approximately (5.243, 3.121).

You are very welcome :)

A line \(y=mx+b\)  intersects the parabola \(y=x^2\) at points \(A\) and \(B\).
The line \(AB\) intersects the y-axis at the point  \(P\).

If \(AP-BP=1\)  then find \(m^2\).

\(\text{Let $A=(x_A,y_A)=(x_A,x_{A}^2) $} \\ \text{Let $B=(x_B,y_B)=(x_B,x_{B}^2) $} \\ \text{Let $P=(x_P,y_P)=(0,y_P) $} \)

\(\begin{array}{|l&rcll|} \hline BP: & m &=& \dfrac{y_P-y_B}{x_P-x_B} \\ & &=& \dfrac{y_P-x_{B}^2}{0-x_B} \\ &\mathbf{ m}&=& \mathbf{\dfrac{x_{B}^2-y_P}{x_B}} \\\\ AP: & m &=& \dfrac{y_A-y_P}{x_A-x_P} \\ & &=& \dfrac{x_{A}^2-y_P}{x_A-0} \\ &\mathbf{m} &=& \mathbf{\dfrac{x_{A}^2-y_P}{x_A}} \\ \hline \mathbf{ m} & = \mathbf{\dfrac{x_{B}^2-y_P}{x_B}}&=& \mathbf{\dfrac{x_{A}^2-y_P}{x_A}} \\ & x_A(x_{B}^2-y_P) &=& x_B(x_{A}^2-y_P)\\ & x_A x_{B}^2-x_Ay_P &=& x_B x_{A}^2-x_By_P \\ & y_P(x_B-x_A) &=& x_B x_{A}^2-x_A x_{B}^2 \\ & y_P(x_B-x_A) &=& -x_Ax_B(x_B-x_A) \\ (1) & \mathbf{y_P} &=& \mathbf{-x_Ax_B} \\ \hline &\mathbf{ m}&=& \mathbf{\dfrac{x_{B}^2-y_P}{x_B}} \quad | \quad \mathbf{y_P=-x_Ax_B} \\ & m &=& \dfrac{x_{B}^2-(-x_Ax_B)}{x_B} \\ & m &=& \dfrac{x_{B}^2+x_Ax_B}{x_B} \\ & m &=& \dfrac{x_{B}^2}{x_B}+\dfrac{x_Ax_B}{x_B} \\ (2) & \mathbf{m} &=& \mathbf{x_A+x_B} \\ \hline \end{array}\)

\(\begin{array}{|rclrcl|} \hline \mathbf{AP-BP} &=& {1} \qquad \text{or} \qquad BP = AP-1 \\\\ \mathbf{AB} &=& \mathbf{AP+BP} \\ AB &=& AP+AP-1 \\ \mathbf{AB} &=& \mathbf{2AP-1} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline AB^2 &=& (x_A -x_B )^2+(y_A -y_B )^2 \\ &=& (x_A -x_B )^2+\left(x_{A}^2 -x_{B}^2 \right)^2 \\ &=& (x_A -x_B )^2+\Big( (x_{A} -x_{B}) (x_{A} +x_{B}) \Big)^2 \\ &=& (x_A -x_B )^2+ (x_{A} -x_{B})^2 (x_{A} +x_{B})^2 \\ &=& (x_A -x_B )^2 \Big(1 +( \underbrace{x_{A} +x_{B}}_{=m})^2\Big) \\ &=& (x_A -x_B )^2 (1 +m^2 ) \\ \mathbf{AB} &=& \mathbf{(x_A -x_B )\sqrt{1 +m^2} } \\\\ AP^2 &=& (x_A -x_P )^2+(y_A -y_P )^2 \\ &=& (x_A -0 )^2+(x_{A}^2 -y_P )^2 \\ &=& x_{A}^2+(x_{A}^2 -y_P )^2 \quad | \quad y_P=-x_Ax_B \\ &=& x_{A}^2+(x_{A}^2 + x_Ax_B )^2 \\ &=& x_{A}^2+\Big(x_A(\underbrace{x_A+x_B}_{=m}) \Big)^2 \\ &=& x_{A}^2+(x_A\times m)^2 \\ &=& x_{A}^2+x_{A}^2m^2 \\ &=& x_{A}^2(1+m^2) \\ \mathbf{AP} &=& \mathbf{x_A\sqrt{1 +m^2} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{AB} &=& \mathbf{2AP-1} \\\\ \mathbf{(x_A -x_B )\sqrt{1 +m^2} } &=& 2\times \mathbf{x_A\sqrt{1 +m^2} } -1 \\ \sqrt{1 +m^2}(x_A-x_B-2x_A) &=& -1 \\ \sqrt{1 +m^2}(2x_A-x_A+x_B) &=& 1 \\ \sqrt{1 +m^2}(\underbrace{x_A+x_B}_{=m}) &=& 1 \\ \sqrt{1 +m^2}\times m &=& 1 \quad \text{square both sides} \\ (1 +m^2)\times m^2 &=& 1 \\ (m^2)^2 + m^2 - 1 &=& 0 \\ m^2 &=& \dfrac{-1\pm \sqrt{1-4*(-1)} }{2} \\\\ m^2 &=& \dfrac{-1 {\color{red}+} \sqrt{1-4*(-1)} }{2} \quad | \quad m^2 > 0 ! \\ \mathbf{m^2} &=& \mathbf{\dfrac{-1 +\sqrt{5} }{2}} \\ \hline \end{array}\)

Ah I see. Thank you Melody!

How do you know if the graph of a fractional function has asymptotes?

Firstly ou look for values that x or y cannot be.  That will take care of the vertical and horizontal asymptotes.

For instance

You cannot divide by 0 so   

   for instance in       \(y=\frac{3}{x+2}\)

x=-2 would be an asymptote, 

also 3 divided by anything cannot be 0 so   y=0 is also an asymptote.

You do have to be careful though because sometimes these values are holes in the graph and not asymptotes at all.

It takes practice to work out the nuances of graphing.

After you think about what a graph might look like enter it into desmos graphing calculator to see if you are right.

That will help you to learn.

Hi Tom  

Thank you!!!!!!

I get it now!!!

Time to start my finals prep...

T-T

Yes, just few steps, not hard. 

I.E, Corollary comes from Theorems (1 theorem could have more than 5 corollaries) (you could say, corollary is derived from theorem but not the verse) 

Corollary is noticed from the theorem , theorem is noticed from a problem (You need a lot of proof for a theorem but no proof or just little for corollary) 

Thank you!!!

But what is your definition of short proof?

Less than 4 steps?

Theorem — a mathematical statement that is proved using rigorous mathematical reasoning.  In a mathematical paper, the term theorem is often reserved for the most important results.

2)Corollary — a result in which the (usually short) proof relies heavily on a given theorem (we often say that “this is a corollary of Theorem A”).

2)Corollary: A statement that follows with little or no proof required from an already proven statement. For example, it is a theorem in geometry that the angles opposite two congruent sides of a triangle are also congruent. A corollary to that statement is that an equilateral triangle is also equiangular.

2z  + i   =  iz +  3

2 (a + bi)  + i   =  i ( a + bi) + 3

2a + 2bi + 1i  =  ai - b + 3

Equating real and complex parts

2a  = 3 - b     ⇒    2a + b  = 3    ⇒  4a + 2b  = 6    (1)

(2b + 1)  =  a ⇒     a - 2b =  1   ⇒    a  - 2b  =  1      (2)

Add (1)  and (2)  and we get that

5a = 7

a = 7/5

And

2(7/5) + b  = 3

14/5 + b = 15/5

b = 15/5 - 14/5

b = 1/5

So

z = a + bi

z  = (7/5)  + (1/5) i

nvm, i understand it now

Source:  https://brilliant.org/problems/marys-square-tiles/?utm_campaign=fb_posts_11152014&utm_medium=social&utm_source=facebook&utm_content=red_blue_tiles

Ah I see. Thank you! I learnt German for two years in school but eventually chose French. I still understand German a bit though because our languages are so similar. 

Find all values of k so that the graphs of x^2+y^2=4+12x+6y and x^2+y^2=k+4x+12y intersect. Enter your answer using interval notation.

\(x^2+y^2=4+12x+6y\;\; and \;\;x^2+y^2=k+4x+12y \\~\\ x^2-12x+y^2-6y=4\\ x^2-12x+36+y^2-6y+9=4+36+9\\ (x-6)^2+(y-3)^2=49\\ \text{circle centre (6,3) radius 7}\\~\\ x^2+y^2=k+4x+12y\\ x^2-4x+y^2-12y=k\\ (x-2)^2+(y-6)^2=k+40\\ \text{circle centre (2,6) radius} =\sqrt{40+k} \)

Now find the distance between the 2 centres.

I got 5

5 is less than 7 so the centre of the variable circle is inside the fixed circle

It the radius of the variable circle is less than 2 then it will be entirely inside the other one and there will be no intersection.

I leave it to you to check everything I have done very carefully and to finish it off neatly.

Please no one provide any more details but asker guest can ask for clarification on details if they want to.

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Coding only:

x^2+y^2=4+12x+6y\;\; and \;\;x^2+y^2=k+4x+12y \\~\\

x^2-12x+y^2-6y=4\\
x^2-12x+36+y^2-6y+9=4+36+9\\
(x-6)^2+(y-3)^2=49\\
\text{circle centre (6,3) radius 7}\\~\\
x^2+y^2=k+4x+12y\\
x^2-4x+y^2-12y=k\\
(x-2)^2+(y-6)^2=k+40\\
\text{circle centre (2,6)  radius} =\sqrt{40+k}

The answer is 22*21*20 = 9240.

The original time  is   D / R  = 200/25  = 8 hrs

So.....he Total Distance is given by 200 + 100T     

Where T is the additional time driving at 100 km / hr

So    Total Distance / Total Time = 40

[200 + 100T ]

___________  =   40      cross-multiply

   8  +  T

200 + 100T  = 40 [ 8 + T]    simplify

200 + 100T  = 320 + 40T

60T  =  120      divide both sides by  60

T  =120 /60   =  2 additional hrs of driving at 100 km.hr  = (2) 100  = 200 additional km 

CoolStuff: Where does your number 892930 appear in the above sequence "leaving 6 in their original order"?

A. Determine the position and velocity functions for the book.

B. Determine the average velocity of the book on the interval [0, 1].

C. Find the instantaneous velocities when t = 0 and t = 1.

D. At what time is the instantaneous velocity of the coin equal to the average velocity of the book found in part B?

E. What is the name of the theorem that says there must be at least one solution to part D?

F. Find the velocity of the book just before it hits the ground.

A.  

The position function is   :  s(t)  =     -16t^2 -10t + 378 

The velocity function isthe derivative of this = s'(t)  =  -32t - 10

B.

The  average velocity on [ 0, 1]  =

[ (-32(1) - 10 )  - (-32(0) - 10 ]            -32

_______________________ =      ______  =    -32 ft/s

           1  - 0                                         1

C. When t  = 0,  the instantaneous velocity  is  -32(0) - 10  =  -10ft/s    [which we would expect ]

     When t = 1, the instantaneous velocity is  -32(1) - 10  =  - 42ft/s

D. I am taking "coin" to mean "book"

We want to know when 

-32  =  - 32t - 10      add 10 to both sides

-22 = -32 t      divide both sides by -32

-22/-32  = t  =  11/16 sec = .6875  sec

E.  Average Value Theorem  [ or, Mean Value Theorem ]

F.  We need to solve this to find the  time it takes to hit the ground

-16t^2 - 10t + 378  = 0

16t^2 + 10t - 378  =  0        divide through by 2

8t^2 + 5t - 189  = 0

Using the quad formula

t =   -5 ±√ [ 5^2 - 4*8*189 ]            -5 ±√ [ 25 + 6048]

      ___________________  =   _________________  ≈  4.558 sec  or  - 5.183 sec

              2 * 8                                         16

Take the positive time  and the instantaneous velocity  when the book hits the ground is :

-32(4.558) - 10   ≈  -155.86 ft/s