Find the vertical and horizontal asymptotes of the function f(x) = (x-2)/(x^2-16).
The vertical asymptote can be found by making the denominator equal to zero and solving for x. x^2-16 = 0 x = 4, -4 Therefore there is a VA at x = 4 and x = -4 The horizontal asymptote can be found by looking at the highest exponent of x, which in this case if x^2. Looking only at that term we find 0/1 = 0. Therefore there is a HA at y = 0.
Complete the missing parts of the following sentence. During photosynthesis water undergoes __________ while carbon dioxide undergoes ________.
The answer to this question is: During photosynthesis water undergoes OXIDATION while carbon dioxide undergoes REDUCTION. This is because water loses electrons while carbon dioxide gains electrons.
What is the tangent line of the function f(x) = 2x^3-4x^2-3x at x = 2?
First we must take the derivative to find the tangent slope at x = 2 f'(x) = 6x^2-8x-3. f'(2) = 6(2)^2-8(2)-3 = 5 Now we must find the y intercept of the tangent line f(2) = 2(2)^3-4(2)^2-3(2) = -6 Using y=mx+b -6 = 5(2)+b b = -16 Therefore the tangent line of the function at x = 2 is y = 5x-16