Need help answering this problem

(a) Local Maximum (x,y) = (0,12)

    Local Minimum (x,y) = (-9,6) (smaller x-value)

    Local Minimum (x,y) = (6,3) (Larger x-value)

(b) Increasing =

    Decreasing =

Step-by-step explanation:

(a)

Local Maximum: The function gives maximum value in small neighborhood.

From the given graph it is clear that the maximum value of the function is 12 at x=0. So,

Local Maximum (x,y) = (0,12)

Local Minimum: The function gives minimum value in small neighborhood.

It is clear that the minimum value of the function is -6 and 3 at x=-9 and x=6. So,

Local Minimum (x,y) = (-9,6) (smaller x-value)

Local Minimum (x,y) = (6,3) (Larger x-value)

(b)

The function is increasing on -9<x<0 and after 6.

Increasing =

The function is decreasing on 0<x<6 and before -9.

Decreasing =

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.