


with that template in mind.
![\bf f(x)=\sqrt[3]{x}\qquad \boxed{-f(x+2)+4}\implies f(x)=\stackrel{A}{-1}\sqrt[3]{\stackrel{B}{1}x\stackrel{C}{+2}}\stackrel{D}{+4}\qquad \\\\\\ f(x)=-\sqrt[3]{x+2}+4](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%3D%5Csqrt%5B3%5D%7Bx%7D%5Cqquad%20%5Cboxed%7B-f%28x%2B2%29%2B4%7D%5Cimplies%20f%28x%29%3D%5Cstackrel%7BA%7D%7B-1%7D%5Csqrt%5B3%5D%7B%5Cstackrel%7BB%7D%7B1%7Dx%5Cstackrel%7BC%7D%7B%2B2%7D%7D%5Cstackrel%7BD%7D%7B%2B4%7D%5Cqquad%20%0A%5C%5C%5C%5C%5C%5C%0Af%28x%29%3D-%5Csqrt%5B3%5D%7Bx%2B2%7D%2B4)
A = -1, reflected over the x-axis
B = 1, C = 2, horizontal shift of C/B or 2, to the left.
D = 4, shifted upwards by 4 units.
check the picture below.
Answer:

Step-by-step explanation:
It's given that
. Because the side
corresponds with the side
, their lengths will be equal. Therefore, we have:

Plugging
, we have:
.
D 25 devided by 2. is the answer
Gradient=rise/run=y2-y1/x2-x1
= 7-3(4)/-4--4(0)
As you can't divide by zero the line's equation is x=-4 so it doesn't have a gradient as such. However, a parallel line would be the y-axis