Answer:
answer below
Step-by-step explanation:
I think you mean "squashed" or "stretched" vertically by a factor of 5.
f(x) is the original parabola equation
g(x) = 5f(x) is vertically stretched by a factor of 5
(x,y) -> (x , 5y)
It's p = 6/(x+a).
Multiply both sides by p to get 6 = p(x+a), then divide both sides by (x+a) to get 6/(x+a) = p.
Check the picture below.
![~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ Q(\stackrel{x_1}{-4}~,~\stackrel{y_1}{9})\qquad U(\stackrel{x_2}{2}~,~\stackrel{y_2}{3}) ~\hfill QU=\sqrt{[ 2- (-4)]^2 + [ 3- 9]^2} \\\\\\ ~\hfill \boxed{QU=\sqrt{72}} \\\\\\ U(\stackrel{x_1}{2}~,~\stackrel{y_1}{3})\qquad A(\stackrel{x_2}{-3}~,~\stackrel{y_2}{-2}) ~\hfill UA=\sqrt{[ -3- 2]^2 + [ -2- 3]^2} \\\\\\ ~\hfill \boxed{UA=\sqrt{50}}](https://tex.z-dn.net/?f=~%5Chfill%20%5Cstackrel%7B%5Ctextit%7B%5Clarge%20distance%20between%202%20points%7D%7D%7Bd%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%7D~%5Chfill~%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20Q%28%5Cstackrel%7Bx_1%7D%7B-4%7D~%2C~%5Cstackrel%7By_1%7D%7B9%7D%29%5Cqquad%20U%28%5Cstackrel%7Bx_2%7D%7B2%7D~%2C~%5Cstackrel%7By_2%7D%7B3%7D%29%20~%5Chfill%20QU%3D%5Csqrt%7B%5B%202-%20%28-4%29%5D%5E2%20%2B%20%5B%203-%209%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20~%5Chfill%20%5Cboxed%7BQU%3D%5Csqrt%7B72%7D%7D%20%5C%5C%5C%5C%5C%5C%20U%28%5Cstackrel%7Bx_1%7D%7B2%7D~%2C~%5Cstackrel%7By_1%7D%7B3%7D%29%5Cqquad%20A%28%5Cstackrel%7Bx_2%7D%7B-3%7D~%2C~%5Cstackrel%7By_2%7D%7B-2%7D%29%20~%5Chfill%20UA%3D%5Csqrt%7B%5B%20-3-%202%5D%5E2%20%2B%20%5B%20-2-%203%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20~%5Chfill%20%5Cboxed%7BUA%3D%5Csqrt%7B50%7D%7D)
![A(\stackrel{x_1}{-3}~,~\stackrel{y_1}{-2})\qquad D(\stackrel{x_2}{-9}~,~\stackrel{y_2}{4}) ~\hfill AD=\sqrt{[ -9- (-3)]^2 + [ 4- (-2)]^2} \\\\\\ ~\hfill \boxed{AD=\sqrt{72}} \\\\\\ D(\stackrel{x_1}{-9}~,~\stackrel{y_1}{4})\qquad Q(\stackrel{x_2}{-4}~,~\stackrel{y_2}{9}) ~\hfill DQ=\sqrt{[ -4- (-9)]^2 + [ 9- 4]^2} \\\\\\ ~\hfill \boxed{DQ=\sqrt{50}}](https://tex.z-dn.net/?f=A%28%5Cstackrel%7Bx_1%7D%7B-3%7D~%2C~%5Cstackrel%7By_1%7D%7B-2%7D%29%5Cqquad%20D%28%5Cstackrel%7Bx_2%7D%7B-9%7D~%2C~%5Cstackrel%7By_2%7D%7B4%7D%29%20~%5Chfill%20AD%3D%5Csqrt%7B%5B%20-9-%20%28-3%29%5D%5E2%20%2B%20%5B%204-%20%28-2%29%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20~%5Chfill%20%5Cboxed%7BAD%3D%5Csqrt%7B72%7D%7D%20%5C%5C%5C%5C%5C%5C%20D%28%5Cstackrel%7Bx_1%7D%7B-9%7D~%2C~%5Cstackrel%7By_1%7D%7B4%7D%29%5Cqquad%20Q%28%5Cstackrel%7Bx_2%7D%7B-4%7D~%2C~%5Cstackrel%7By_2%7D%7B9%7D%29%20~%5Chfill%20DQ%3D%5Csqrt%7B%5B%20-4-%20%28-9%29%5D%5E2%20%2B%20%5B%209-%204%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20~%5Chfill%20%5Cboxed%7BDQ%3D%5Csqrt%7B50%7D%7D)
now, let's take a peek at that above, DQ = UA and QU = AD, so opposite sides of the polygon are equal.
Now, let's check the slopes of DQ and QU
![D(\stackrel{x_1}{-9}~,~\stackrel{y_1}{4})\qquad Q(\stackrel{x_2}{-4}~,~\stackrel{y_2}{9}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{9}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{-4}-\underset{x_1}{(-9)}}} \implies \cfrac{9 -4}{-4 +9}\implies \cfrac{5}{5}\implies \cfrac{1}{1}\implies 1 \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=D%28%5Cstackrel%7Bx_1%7D%7B-9%7D~%2C~%5Cstackrel%7By_1%7D%7B4%7D%29%5Cqquad%20Q%28%5Cstackrel%7Bx_2%7D%7B-4%7D~%2C~%5Cstackrel%7By_2%7D%7B9%7D%29%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7Bslope%7D%7Bm%7D%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%20%7B%5Cstackrel%7By_2%7D%7B9%7D-%5Cstackrel%7By1%7D%7B4%7D%7D%7D%7B%5Cunderset%7Brun%7D%20%7B%5Cunderset%7Bx_2%7D%7B-4%7D-%5Cunderset%7Bx_1%7D%7B%28-9%29%7D%7D%7D%20%5Cimplies%20%5Ccfrac%7B9%20-4%7D%7B-4%20%2B9%7D%5Cimplies%20%5Ccfrac%7B5%7D%7B5%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B1%7D%5Cimplies%201%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![Q(\stackrel{x_1}{-4}~,~\stackrel{y_1}{9})\qquad U(\stackrel{x_2}{2}~,~\stackrel{y_2}{3}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{3}-\stackrel{y1}{9}}}{\underset{run} {\underset{x_2}{2}-\underset{x_1}{(-4)}}} \implies \cfrac{3 -9}{2 +4}\implies \cfrac{-6}{6}\implies \cfrac{-1}{1}\implies -1](https://tex.z-dn.net/?f=Q%28%5Cstackrel%7Bx_1%7D%7B-4%7D~%2C~%5Cstackrel%7By_1%7D%7B9%7D%29%5Cqquad%20U%28%5Cstackrel%7Bx_2%7D%7B2%7D~%2C~%5Cstackrel%7By_2%7D%7B3%7D%29%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7Bslope%7D%7Bm%7D%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%20%7B%5Cstackrel%7By_2%7D%7B3%7D-%5Cstackrel%7By1%7D%7B9%7D%7D%7D%7B%5Cunderset%7Brun%7D%20%7B%5Cunderset%7Bx_2%7D%7B2%7D-%5Cunderset%7Bx_1%7D%7B%28-4%29%7D%7D%7D%20%5Cimplies%20%5Ccfrac%7B3%20-9%7D%7B2%20%2B4%7D%5Cimplies%20%5Ccfrac%7B-6%7D%7B6%7D%5Cimplies%20%5Ccfrac%7B-1%7D%7B1%7D%5Cimplies%20-1)
keeping in mind that perpendicular lines have negative reciprocal slopes, let's notice that the reciprocal of 1/1 is just 1/1 and the negative of that is just -1/1 or -1, so QU has a slope that is really just the negative reciprocal of DQ, those two lines are perpendicular, thus making a 90° angle, and their congruent opposite sides will also make a 90°, that makes QUAD hmmm yeap, you guessed it.
Answer:
- 6
Step-by-step explanation:
The average rate of change of f(x) in the closed interval [ a, b ] is
![\frac{f(b)-f(a)}{b-a}](https://tex.z-dn.net/?f=%5Cfrac%7Bf%28b%29-f%28a%29%7D%7Bb-a%7D)
here [ a, b ] = [ - 7, - 1 ]
f(b) = f(- 1) = (- 1)² + 2(- 1) - 8 = 1 - 2 - 8 = - 9
f(a) = f(- 7) = (- 7)² + 2(- 7) - 8 = 49 - 14 - 8 = 27
Hence
average rate of change =
=
= - 6
Answer:
--1/2 or -0.5
Step-by-step explanation: