Answer: equation of the tangent plane is z = 1
Step-by-step explanation:
Given equation
z = e^(-x²-y²) at point (0,0,1)
now let z = f(x,y)
Δf(x,y) = [ fx, fy ]
= (-2xe^(-x²-y²)), (-2ye^(-x²-y²))
now
Δf (0,0) = [ 0, 0 ] = [ a, b ]
equation of the tangent plane therefore will be
z - z₀ = a(x-x₀) + b(y-y₀)
z - 1 = 0(x-0) + 0(y-0)
z - 1 = 0 + 0
z = 1
Therefore equation of the tangent plane is z = 1
Y = mx + b
plug in the numbers.
y = 4/3 + 2
Look carefully at the first pair: (−3, 9), (−3, −5) Note that x does not change, tho' y does. This is how we recognize a vertical line (whose slope is undefined). The equation of this vertical line is x = -3.
Looking at the second pair: from (3,4) to (5,6), x increases by 2 and y by 2; thus, the slope is m = rise/run = 2/2 = 1.
Third pair: as was the case with the first pair, x does not change here, and thus the equation of this (vertical) line is x=0 (which is the y-axis). The slope is undefined.
Keep change flip. You keep the first fraction the same, change the sign to multiplication and flip the second fraction. Example: 3/4 divided by 1/2 would be worked as 3/4 times 2/1
