Answer:
The two horiz. tang. lines here are y = -3 and y = 192.
Step-by-step explanation:
Remember that the slope of a tangent line to the graph of a function is given by the derivative of that function. Thus, we find f '(x):
f '(x) = x^2 + 6x - 16. This is the formula for the slope. We set this = to 0 and determine for which x values the tangent line is horizontal:
f '(x) = x^2 + 6x - 16 = 0. Use the quadratic formula to determine the roots here: a = 1; b = 6 and c = -16: the discriminant is b^2-4ac, or 36-4(1)(-16), which has the value 100; thus, the roots are:
-6 plus or minus √100
x = ----------------------------------- = 2 and -8.
2
Evaluating y = x^3/3+3x^2-16x+9 at x = 2 results in y = -3. So one point of tangency is (2, -3). Remembering that the tangent lines in this problem are horizontal, we need only the y-coefficient of (2, -3) to represent this first tangent line: it is y = -3.
Similarly, find the y-coeff. of the other tangent line, which is tangent to the curve at x = -8. The value of x^3/3+3x^2-16x+9 at x = -8 is 192, and so the equation of the 2nd tangent line is y=192 (the slope is zero).
M = mile
.55m + 1.45 [less than or equal to] 35
subtract 1.45
.55m = 33.55
divide by .55
m = 61 miles or less
Answer:
56
Step-by-step explanation:
Since 3 is between -6 and 9 we use the first definition
f(3) = 6 x^2 +2
= 6(3)^2+2
= 6(9) +2
= 54+2
= 56
30:18,
31:23
32:28
33:33
It will be three months before they have the same amount of CDs
3 tens, 2 ones, 1 tenth, 5 hundredths
