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:
6 + 6 = 12 × 10 = ?? = 132 that would be just guess the other
Answer:
slope is 2 and y-intercept is (0, 4)
Step-by-step explanation:
Answer:
-75/48
Step-by-step explanation:
multiply 5 by 15
and 6 by 8
5x15 is 75 , put it in numerator
6x8 is 48 , denominator
then the answer should have the negative sign
Given:
Slope = -3/5
y - intercept = (0 , 5)
The answer is y = -3/5 + 5
Note:
The x = 0 and the y = 5 . We are looking for y- intercept so we use 5 instead of 0.
Since the formula for slope intercept form is y = mx + b, we use option number 4 as our equation. for option number 1, it did not include the negative sign that was given in the slope so we opt to choose option 1.