to get the equation of any straight line, all we need is two points from it, Check the picture below, let's use those two.
![(\stackrel{x_1}{-5}~,~\stackrel{y_1}{-5})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{(-5)}}}{\underset{run} {\underset{x_2}{0}-\underset{x_1}{(-5)}}}\implies \cfrac{1+5}{0+5}\implies \cfrac{6}{5}](https://tex.z-dn.net/?f=%28%5Cstackrel%7Bx_1%7D%7B-5%7D~%2C~%5Cstackrel%7By_1%7D%7B-5%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B0%7D~%2C~%5Cstackrel%7By_2%7D%7B1%7D%29%20~%5Chfill%20%5Cstackrel%7Bslope%7D%7Bm%7D%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%20%7B%5Cstackrel%7By_2%7D%7B1%7D-%5Cstackrel%7By1%7D%7B%28-5%29%7D%7D%7D%7B%5Cunderset%7Brun%7D%20%7B%5Cunderset%7Bx_2%7D%7B0%7D-%5Cunderset%7Bx_1%7D%7B%28-5%29%7D%7D%7D%5Cimplies%20%5Ccfrac%7B1%2B5%7D%7B0%2B5%7D%5Cimplies%20%5Ccfrac%7B6%7D%7B5%7D)
![\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-5)}=\stackrel{m}{\cfrac{6}{5}}(x-\stackrel{x_1}{(-5)}) \\\\\\ y+5=\cfrac{6}{5}(x+5)\implies y+5=\cfrac{6}{5}x+6\implies y=\cfrac{6}{5}x+1](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20%5Ctextit%7Bpoint-slope%20form%7D%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y-y_1%3Dm%28x-x_1%29%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%5Cimplies%20y-%5Cstackrel%7By_1%7D%7B%28-5%29%7D%3D%5Cstackrel%7Bm%7D%7B%5Ccfrac%7B6%7D%7B5%7D%7D%28x-%5Cstackrel%7Bx_1%7D%7B%28-5%29%7D%29%20%5C%5C%5C%5C%5C%5C%20y%2B5%3D%5Ccfrac%7B6%7D%7B5%7D%28x%2B5%29%5Cimplies%20y%2B5%3D%5Ccfrac%7B6%7D%7B5%7Dx%2B6%5Cimplies%20y%3D%5Ccfrac%7B6%7D%7B5%7Dx%2B1)
Answer: D
Step-by-step explanation:
Answer:
$5,552.26
Step-by-step explanation:
First, set up an equation using exponential decay
Let y represent the cost and let x represent the number of years
y = 15,000(0.78)^x
Plug in 4 as x, and simplify:
y = 15,000(0.78)^4
y = 5,552.26
So, in 4 years, the car will be $5,552.26
Answer:
y = -2(x -2)^2 + 11
Step-by-step explanation:
It works well to factor the leading coefficient from the first two terms.
... y = -2(x^2 -4x) +3
Now we want to add the square of half the x-coefficient inside parentheses, and subtract the equivalent quantity outside parentheses.
... y = -2(x^2 -4x +4) +3 - (-2·4)
... y = -2(x -2)^2 +11 . . . . . . . . simplify
_____
The form given in the problem statement is called "vertex form," where the vertex of the parabola is (h, k). A graph shows us the vertex is (2, 11), so we can write the function immediately as ...
... y = -2(x -2)^2 +11