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Since  x = -3  and  x = 2   are vertical asymptotes, then we might guess that the denominator might be of the form...

a(x + 3) (x - 2)

Since  (0,0)  is on the graph, then the numerator must be of the form.... bx

And the points  (3,1)  and (-2, 1)   are on the graph

So...we have that

        b (3)                                  3b 

____________   =    1    ⇒    ________ =    1   ⇒   3b  = 6a  ⇒  b = 2a

a (3 + 3)(3 - 2 )                          6a

And

     b(-2)                                        -2b

______________  =     1   ⇒     ________  =  1  ⇒   -2b = -4a   ⇒  -2(2a)  = -4a ⇒ a  = 1 

a(-2 + 3) (-2 - 2)                         a(1)(-4)

So  p   = bx =    2(a)x  =  2(1)x  =   2x

And q  =  1 (x + 3) (x - 2)  =  (x + 3) (x - 2)

So  p(-1 )               2(-1)                       -2                   -1            1

      ____   =     ____________   =   ________  =     ___  =     ___

      q (-1)         (-1 + 3) (-1 - 2)           (2) (-3)              -3            3

a=90000; c=  8*(a-2)^5-a^2+14*a-24; if(c%5==0, goto3, goto4);printc, a; a++;if(a<110000, goto1, 0)

The largest value of n < 100,000 is n =99,997

2)  This one is tricky but here's my attempt...

I started by making this list:

By now we can notice something...

Between  4  and  9 ,  there are  4  numbers and  4/2  =  2

Between  9  and  16  there are  6  numbers and  6/2  =  3

So the number of 3's   =  1 + 2 + 3   =   6

We can backtrack the previous steps to get...

the number of 3's   =   \(1+\frac{4}{2}+\frac62\ =\ 1+\frac{9-4-1}{2}+\frac{16-9-1}{2}\ =\ 1+\frac{3^2-(3-1)^2-1}{2}+\frac{(3+1)^2-3^2-1}{2}\)

So in general,

the number of  n's   =   \(1+\frac{n^2-(n-1)^2-1}{2}+\frac{(n+1)^2-n^2-1}{2}\)     which simplifies to...

the number of n's   =   2n

This agrees with the list from earlier!!!

Except the number of  45's  is only  27  because the list ends at  b(2007)

\(\sum\limits_{p=1}^{2007}b(p)\ =\ b(1)+b(2)+b(3)+b(4)+b(5)+b(6)+b(7)+\dots+b(2006)+b(2007)\\~\\ \sum\limits_{p=1}^{2007}b(p)\ =\ 1+1+2+2+2+2+3+\dots+45+45\\~\\ \sum\limits_{p=1}^{2007}b(p)\ =\ 2(1)+4(2)+6(3)+\dots+88(44)+27(45)\\~\\ \sum\limits_{p=1}^{2007}b(p)\ =\ \sum\limits_{k=1}^{44}(2k)(k)\quad+\ 27(45)\\~\\ \sum\limits_{p=1}^{2007}b(p)\ =\ \sum\limits_{k=1}^{44}(2k)(k)\quad+\ 27(45)\\~\\ \sum\limits_{p=1}^{2007}b(p)\ =\ 2\sum\limits_{k=1}^{44}k^2\quad+\ 27(45)\\~\\ \sum\limits_{p=1}^{2007}b(p)\ =\ 2(\frac{44(44+1)(2(44)+1)}{6})\quad+\ 27(45)\\~\\ \sum\limits_{p=1}^{2007}b(p)\ =\ 58740\quad+\ 1215\\~\\ \sum\limits_{p=1}^{2007}b(p)\ =\ 59955 \)

P.S. There definitely might be a better way to do it...

Find a+b+c if the graph of the equation y=ax^2+bx+c is a parabola with vertex (5,3), vertical axis of symmetry, and contains the point (2,0).

(2.0)  means that x  =  2  is a root

The x  value of the other root is  given by

[ 2 + x ]

_______   =        5               multiply through by 2

     2

2 + x  =  10             subtract  2 from both sides

x  = 8

So.....the other root is   (8, 0)

So  we have that

y  =  a (x - 2) (x - 8)          and since    (5,3)  is on the graph, we can find "a"  as

3  =  a ( 5 - 2) (5 - 8)

3  = a (3)(-3)

3 = - 9a

a =  -1/3

So we have that

y  = (-1/3) (x - 2) (x - 8)

y  = (-1/3) (x^2 - 10x + 16)

y = (-1/3)x^2  + (10/3)x  - (16/3)

So     a  +  b   +  c   =   (-1/3)  + (10/3)  - (16/3)   =   -7/3

the answer was 54 for those who are wondering...

See hectictar's answer here : https://web2.0calc.com/questions/help-urgent_28#r1

ok its fine

What? She does have a "pinky" avatar...

1)  ( Assuming you meant compute f(5,5) )

k  =  (6x + 12)(x - 8)

                                          Multiply out the right side of the equation.

k  =  6x² - 48x + 12x - 96

                                          Combine like terms.

k  =  6x² - 36x - 96

In an equation of the form   k  =  ax² + bx + c   with   a > 0,

the least possible value of   k   occurs at   \(x\ =\ \text-\frac{b}{2a}\)

So...

In an equation of the form   k  =  6x² - 36x - 96   with   6 > 0,

the least possible value of   k   occurs at   \(x\ =\ \text-\frac{(\text-36)}{2(6)}\ =\ \frac{36}{12}\ =\ 3\)

When  x = 3,     k   =   6(3)² - 36(3) - 96   =   6(9) - 36(3) - 96   =   54 - 108 - 96   =   -150

x = 3  and  y = 2

Anyone?

Well, some values are x=0, y=2, and x=3, y=0. Slope is -2/3, so perpendicular is 3/2.

Yup... I am. Forgot to login.

Oh wow, that's just rude XD

Here, symmetry is used to tackle questions 1 and 2:

1. Both the constraint and the function to be maximised are symmetrical in x, y and z, so we should expect them all to have the same value at the maximum point. Hence let y=x and z=x so the constraint becomes: 3x² = 1 That is x = 1/√3 so x+y+z = 3/√3 = √3

2. Similarly, symmetry suggests a and b should take the same value.   Each term in the function to be minimised is of the form k/x where k is a constant.  The rate of change of a term like this with respect to x is -k/x².  We'd like the rate of change of each term in the function to contribute equally to the minimum value, so we want  -1/a² = -4/c²  and -1/a² = -16/d² so that c = 2a and d = 4a (we already have b = a ) Putting these values in the constraint equation we get  a + a + 2a + 4a = 1  or a = 1/8 Hence b = 1/8,  c = 2/8 and d = 4/8, so 1/a + 1/b + 4/c + 16/d = 64 the minimum value.

Well.....if we can just use a "slight" bit of trig.....

The diagonals of a parallelogram bisect each other.....so.........these diagonals form four cogruent triangles

Each of these triangles will have two sides with lengths of 4 and 5 and an included angle between these sides of 45°

And the area of one of these triangles  =  (1/2) (product of these sides) * sine of the included angle

So.....the area of one of the triangles  =  (1/2) (4 * 5) * sin (45°)

[ sin (45°  =    √2 / 2 ]....so we have

(1/2) ( 4 * 5) ( √2/2)

(1/4) (20) * √2

And since we havefour of these triangles....the total area is

4 (1/4) * (20) * √2  =

20√2  units^2

Thanks, heureka.....that's a pretty neat method for solving this one......

Are you DurgneyDabs? I can't really tell- try logging back in.

For the graph of a certain quadratic y = ax^2 + bx + c,
the vertex of the parabola is (3,7) and one of the x-intercepts is (-2,0).
What is the x-coordinate of the other x-intercept?

\(\text{Let $x-vertex = x_v = 3 $} \\ \text{Let $ x-intercept_1 = x_1 = -2 $} \\ \text{Let $ x-intercept_2 = x_2 =\ ? $} \)

\(\begin{array}{|rcll|} \hline \dfrac{x_1+x_2}{2} &=& x_v \\\\ \dfrac{-2+x_2}{2} &=& 3 \quad | \quad \cdot 2 \\\\ -2+x_2 &=& 6 \quad | \quad + 2 \\\\ \mathbf{ x_2 } &=& \mathbf{8} \\ \hline \end{array} \)

The x-coordinate of the other x-intercept is \(\mathbf{ 8 }\)

\( \prod_{i=1}^{k} a_{i}= {a_1}^{\frac{{2}^{k+1}-1+\frac{{(-1)}^{k+1}-1}{2}}{3}}={2}^{\frac{{2}^{k+1}-1+\frac{{(-1)}^{k+1}-1}{2}}{57}}\)

3.
The sequence \((a_n)\) is defined recursively by \(a_0=1,\ a_1=\sqrt[19]{2}\), and \(a_n=a_{n-1}a_{n-2}^2\) for \(n\geq 2\).
What is the smallest positive integer k such that the product \(a_1a_2\cdots a_k\) is an integer?

\(\begin{array}{|l|r|rcl|l|} \hline & k & \text{product } a_1a_2\cdots a_k \\ \hline a_2 = a_1a_0^2=a_1^1 & 2 & a_1a_2 &=& a_1^2 & =2^{\frac{2}{19}} & = 1.07569058622 \\ \hline a_3 = a_2a_1^2=a_1^1a_1^2=a_1^3 & 3 & a_1a_2a_3 &=& a_1^5 & =2^{\frac{5}{19}} & = 1.20010271958 \\ \hline a_4 = a_3a_2^2=a_1^3a_1^2=a_1^5& 4 & a_1a_2\cdots a_4&=& a_1^{10} & =2^{\frac{10}{19}} & = 1.49375896165 \\ \hline a_{5} = a_4a_3^2=a_1^{5}a_1^{6}=a_1^{11} & 5 & a_1a_2\cdots a_5 &=& a_1^{21} & =2^{\frac{21}{19}} & = 2.15138117244 \\ \hline a_{6} = a_5a_4^2=a_1^{11}a_1^{10}=a_1^{21} & 6 & a_1a_2\cdots a_6 &=& a_1^{42} & =2^{\frac{42}{19}} & = 4.62844094913 \\ \hline a_{7} = a_6a_5^2=a_1^{21}a_1^{33}=a_1^{43} & 7 & a_1a_2\cdots a_7 &=& a_1^{85} & =2^{\frac{85}{19}} & = 22.2184182818 \\ \hline a_{8} = a_7a_6^2=a_1^{43}a_1^{42}=a_1^{85} & 8 & a_1a_2\cdots a_8 &=& a_1^{170} & =2^{\frac{170}{19}} & = 493.658110947 \\ \hline a_{9} = a_8a_7^2=a_1^{85}a_1^{86}=a_1^{171} & 9 & a_1a_2\cdots a_9 &=& a_1^{341} & =2^{\frac{341}{19}} & = 252752.952805 \\ \hline a_{10} = a_9a_8^2=a_1^{171}a_1^{170}=a_1^{341} & 10 & a_1a_2\cdots a_{10} &=& a_1^{682} & =2^{\frac{682}{19}} & = 2^{35.8947368421} \\ \hline a_{11} = a_{10}a_{9}^2=a_1^{341}a_1^{342}=a_1^{683} & 11 & a_1a_2\cdots a_{11} &=& a_1^{1365} & =2^{\frac{1365}{19}} & = 2^{71.8421052632} \\ \hline a_{12} = a_{11}a_{10}^2=a_1^{683}a_1^{682}=a_1^{1365} & 12 & a_1a_2\cdots a_{12} &=& a_1^{2730} & =2^{\frac{2730}{19}} & = 2^{143.684210526} \\ \hline a_{13} = a_{12}a_{11}^2=a_1^{1365}a_1^{1366}=a_1^{2731} & 13 & a_1a_2\cdots a_{13} &=& a_1^{5461} & =2^{\frac{5461}{19}} & = 2^{287.421052632} \\ \hline a_{14} = a_{13}a_{12}^2=a_1^{2731}a_1^{2730}=a_1^{5461} & 14 & a_1a_2\cdots a_{14} &=& a_1^{10922} & =2^{\frac{10922}{19}} & = 2^{574.842105263} \\ \hline a_{15} = a_{14}a_{13}^2=a_1^{5461}a_1^{5462}=a_1^{10923} & 15 & a_1a_2\cdots a_{15} &=& a_1^{21845} & =2^{\frac{21845}{19}} & = 2^{1,149.73684211} \\ \hline a_{16} = a_{15}a_{14}^2=a_1^{10923}a_1^{10922}=a_1^{21845} & 16 & a_1a_2\cdots a_{16} &=& a_1^{43690} & =2^{\frac{43690}{19}} & = 2^{2,299.47368421} \\ \hline a_{17} = a_{16}a_{15}^2=a_1^{21845}a_1^{21846}=a_1^{43691} & \mathbf{17} & a_1a_2\cdots a_{17} &=& a_1^{87381} & =2^{\frac{87381}{19}} & \mathbf{= 2^{4599}}\\ & & && & & \text{the product $a_1a_2\cdots a_{17}$} \\ & & && & & \text{with $k=17$ is an integer} \\ \hline \end{array}\)

The smallest positive integer \(k\) such that the product \(a_1a_2\cdots a_k\) is an integer is \(\mathbf{17}\)