Step 1: Find f'(x):
f'(x) = -6x^2 + 6x
Step 2: Evaluate f'(2) to find the slope of the tangent line at x=2:
f'(2) = -6(2)^2 + 6(2) = -24 + 12 = -12
Step 3: Find f(2), so you have a point on y=f(x):
f(2) = -2·(2)^3 + 3·(2)^2 = -16 + 12 = -4
So, you have the point (2,-4) and the slope of -12.
Step 4: Find the equation of your tangent line:
Using point-slope form you'd have: y + 4 = -12 (x - 2)
That is the equation of the tangent line.
If your teacher is picky and wants slope-intercept, solve that for y to get:
y = -12 x + 20
Answer:
yes, there are an infinite number of solutions
Step-by-step explanation:
The attached graph shows values of A and B (x and y) that can make this equation be true. Any of the points on any of the curves will satisfy the equation. (The repetition continues indefinitely in all directions.)
Answer:
R = 6S/T
Step-by-step explanation:
R ∝ S/ T
》R = kS/ T
》k = RT/ S
If R = 8 when S = 4 and T = 3
then, k = (8 × 3)/ 4
k = 6
∴ R = (6 × S)/ T
1/9 on a calculator is 0.11111111
.3455 is the correct answer.
