Two planes that intersect are simply called a plane to plane intersection. When they intersect, the intersection point is simply called a line.
Therefore, Plane R and Plane S form a line when they intersect.
Not sure what form you want this in:
Simplified this is 4x+4y
Factorised this is 4(x+y)
Answer:
I think it's -14
Step-by-step explanation:
x = 0 so the x won't be included but
-2 • x = -2x
-12 • x = -12x
-2x + (-)12x = -14
hope this helps :)
A generic point on the graph of the curve has coordinates
![(x, 4x^2+1)](https://tex.z-dn.net/?f=%28x%2C%204x%5E2%2B1%29)
The derivative gives us the slope of the tangent line at a given point:
![f(x) = 4x^2+1 \implies f'(x) = 8x](https://tex.z-dn.net/?f=f%28x%29%20%3D%204x%5E2%2B1%20%5Cimplies%20f%27%28x%29%20%3D%208x)
Let k be a generic x-coordinate. The tangent line to the curve at this point will pass through
and have slope ![8k](https://tex.z-dn.net/?f=8k)
So, we can write its equation using the point-slope formula: a line with slope m passing through
has equation
![y-y_0 = m(x-x_0)](https://tex.z-dn.net/?f=y-y_0%20%3D%20m%28x-x_0%29)
In this case,
and
, so the equation becomes
![y-4k^2-1 = 8k(x-k)](https://tex.z-dn.net/?f=y-4k%5E2-1%20%3D%208k%28x-k%29)
We can rewrite the equation as follows:
![y-4k^2-1 = 8k(x-k) \iff y = 8kx - 8k^2+4k^2+1 \iff y = 8kx-4k^2+1](https://tex.z-dn.net/?f=y-4k%5E2-1%20%3D%208k%28x-k%29%20%5Ciff%20y%20%3D%208kx%20-%208k%5E2%2B4k%5E2%2B1%20%5Ciff%20y%20%3D%208kx-4k%5E2%2B1)
We know that this function must give 0 when evaluated at x=0:
![f(x) = 8kx-4k^2+1 \implies f(0) = -4k^2+1 = 8 \iff -4k^2 = 7 \iff k^2 = -\dfrac{7}{4}](https://tex.z-dn.net/?f=f%28x%29%20%3D%208kx-4k%5E2%2B1%20%5Cimplies%20f%280%29%20%3D%20-4k%5E2%2B1%20%3D%208%20%5Ciff%20-4k%5E2%20%3D%207%20%5Ciff%20k%5E2%20%3D%20-%5Cdfrac%7B7%7D%7B4%7D)
This equation has no real solution, so the problem looks impossible.