Y = mx + b
slope(m) = 4
(-3,-1)...x = -3 and y = -1
now we sub...we r looking for b, the y int.
-1 = 4(-3) + b
-1 = -12 + b
-1 + 12 = b
11 = b
so ur equation is : y = 4x + 11
Answer: Choice C
h(x) = -x^4 + 2x^3 + 3x^2 + 4x + 5
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Explanation:
When reflecting the function f(x) over the y axis, we replace every x with -x and simplify like so
f(x) = -x^4 - 2x^3 + 3x^2 - 4x + 5
f(-x) = -(-x)^4 - 2(-x)^3 + 3(-x)^2 - 4(-x) + 5
f(-x) = -x^4 + 2x^3 + 3x^2 + 4x + 5
h(x) = -x^4 + 2x^3 + 3x^2 + 4x + 5
Note the sign changes that occur for the terms that have odd exponents (the terms -2x^3 and -4x become +2x^3 and +4x); while the even exponent terms keep the same sign.
The reason why we replace every x with -x is because of the examples mentioned below
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Examples:
The point (1,2) moves to (-1,2) after a y axis reflection
Similarly, (-5,7) moves to (5,7) after a y axis reflection.
As you can see, the y coordinate stays the same but the x coordinate flips in sign from negative to positive or vice versa. This is the direct reason for the replacement of every x with -x.
Answer:2 11/24
Step-by-step explanation:
The problem can be solved step by step, if we know certain basic rules of summation. Following rules assume summation limits are identical.
![\sum{a+b}=\sum{a}+\sum{b}](https://tex.z-dn.net/?f=%5Csum%7Ba%2Bb%7D%3D%5Csum%7Ba%7D%2B%5Csum%7Bb%7D)
![\sum{kx}=k\sum{x}](https://tex.z-dn.net/?f=%5Csum%7Bkx%7D%3Dk%5Csum%7Bx%7D)
![\sum_{r=1}^n{1}=n](https://tex.z-dn.net/?f=%5Csum_%7Br%3D1%7D%5En%7B1%7D%3Dn)
![\sum_{r=1}^n{r}=n(n+1)/2](https://tex.z-dn.net/?f=%5Csum_%7Br%3D1%7D%5En%7Br%7D%3Dn%28n%2B1%29%2F2)
Armed with the above rules, we can split up the summation into simple terms:
![\sum_{r=1}^n{40r-21n+8}=n](https://tex.z-dn.net/?f=%5Csum_%7Br%3D1%7D%5En%7B40r-21n%2B8%7D%3Dn)
![=40\sum_{r=1}^n{r}-21n\sum_{r=1}^n{1}+8\sum_{r=1}^n{1}](https://tex.z-dn.net/?f=%3D40%5Csum_%7Br%3D1%7D%5En%7Br%7D-21n%5Csum_%7Br%3D1%7D%5En%7B1%7D%2B8%5Csum_%7Br%3D1%7D%5En%7B1%7D)
![=40\frac{n(n+1)}{2}-21n^2+8n](https://tex.z-dn.net/?f=%3D40%5Cfrac%7Bn%28n%2B1%29%7D%7B2%7D-21n%5E2%2B8n)
![=20n(n+1)-21n^2+8n](https://tex.z-dn.net/?f=%3D20n%28n%2B1%29-21n%5E2%2B8n)
![=28n-n^2](https://tex.z-dn.net/?f=%3D28n-n%5E2)
=> (a)
f(x)=28n-n^2=> f'(x)=28-2n
=> at f'(x)=0 => x=14
Since f''(x)=-2 <0 therefore f(14) is a maximum
(b)
f(x) is a maximum when n=14
(c)
the maximum value of f(x) is f(14)=196