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:
No, a porportional relationship is not shown
Step-by-step explanation:
Up until 4(x) and 10(y), the relationship is porportional. Then, when x increases by 2, y also only increases by 2. Y would need to increase by 5.
Here u go the answer is 4/3
12.195+9.13=21.325 is the answer
Answer:
7y + 2x + 34 = 0
Step-by-step explanation:
The gradient for perpendicular, m = -2/7
The formular equation, y - y1 = m(x - x1)
The equation of line through (4, -6) is:
y - (-6) = -2/7(x - 4)
y + 6 = -2x/7 + 8/7
7y + 42 + 2x - 8 = 0
7y + 2x + 34 = 0
