CPhill

Go ahead and post it.....the distance from the horizontal endpoint to where  =  46   ???

You're close, CU......this actually is a parabola that opens to the right

We can find the y intercepts by setting  x  = 0  and solving

y^2 - 4y  - 1   = 0

y^2 - 4y  =   1        complete the square on y

y^2 - 4y + 4  =   1 + 4

(y - 2)^2  =  5      take both roots

y - 2 = ±√5

y  =  2 ±√5

So.....there are two y intercepts  as shown by CU's  graph....!!!!

I'm assuming that the  distance from the center to the focus  =  46   =  c

116/2  =  a  = 58

In an ellipse

b^2 + c^2  = a^2        ....so

b^2  = a^2  - c^2

b^2  = 58^2  - 46^2

b^2  = 3136  - 2116

b = √[ 3136 - 2116 ]  = √1020 = 2√255

OOPS!!!....thanks, EP!!!

OK....good deal   !!!

5x - 17  = -1x  + 7       add 17, 1x to both sides

6x  = 24   divide both sides by 6

x  = 4

QR  splits the sides PS and PT proportionally.....

So

PQ / PR  =  PS / PT

PQ / PR  = [ PQ + QS] / PT

8/ PR  =  [ 8+ 6 ]  / 10.5

8 /PR  = 14/10.5

PR / 8  =  10.5 / 14

PR  =  8 (10.5) /14  =  6

f(0)  =  0^(0+1) * (0 + 2)^(0 + 3)  =  0^1  * 2^3  =   0

f(-1) =  (-1)^(-1 + 1) * (-1 + 2)^(-1 + 3)  = (-1)^0 * (1)^(2)  = 1 * 1  = 1

f(-2)  =  (-2)^(-2 + 1) * (-2 +2)^(-2 + 3)  =  (-2)^(-1)* (0)^(1)  =  0

f(-3)  =  (-3)^(-3 + 1) * (-3 + 2)^(-3 + 3)  = (-3)^(-2) * (-1)^0  =  (1/9)*(1)  = 1/9

The sum of these  =   1 1/9    =   10/9

a *  b   =  a  + a/b   =   12 + 12/3   =  12 + 4   =   16

f(x)  =  g (g(x))

g (g (x))    means that we sub g(x) into itself....so we have that

g(g(x) ) = 2 (2x - 1) - 1  =

4x - 2 - 1  =

4x - 3

So....since this is  f(x)

Then f(3)  =

4(3)  - 3  =

12 - 3  =

9