CPhill

x^2 - 8x + 8 + c

The x coordinate of the vertex is   8/ (2*1)  = 4

So....we  need y to be 1  when x  = 4

1  = (4)^2 - 8(4) + 8 + c

1 = 16 - 32 + 8 + c

1 = -8 + c

c  =  9

Here's the graph :

Note that  if c is larger than 9, the graph is  shited above y =1....so 1 will not be in the range

Simplify as

sqrt [ -3x^2 - x + 6  ]

-3x^2 - x + 6 =  0

3x^2 + x - 6   = 0

x  =   [ -1 - sqrt 73 ] / 6   ≈ -1.59     and  x = [ 1 +sqrt 73  ] / 6 ≈ 1.26

The root will be real when x =  [  (-1- sqrt 73)/6  , (-1 +sqrt 73)/6 ]  = the domain

Second one

Let  O be the center of the circle

Let  M be the mid-point  of UV

Let d be the distance from O to M

We can form a right triangle such that

d^2 + (UV/2)^2  = r^2

d^2 + 19^2  =  r^2      (1)

Let the midpoint of YZ = N

Let the distance from O to N =  d + 4

We can  form another right triangle such that

(d + 4)^2 + (YZ/2)^2  = r^2 

(d + 4)^2 + 11^2  = r^2      (2)

Equate (1) , (2)

d^2 + 19^2   =  (d + 4)^2  + 11^2

d^2 + 361  = d^2 + 8d + 16 + 121

8d =  361  - 16 - 121

8d =  224

d =  224/8  = 28

28^2 +19^2   =r^2

1145 =  r^2

And we can form a third right triangle such that

(d + 2)^2 + (WX / 2)^2  = r^2

(28 + 2)^2  + WX^2 / 4  = 1145

30^2  + WX^ 2 /4 = 1145

900 + WX^2/4 = 1145

WX^2 / 4  = 1145 - 900

WX^2 / 4  = 245

WX^2   =  245 * 4 = 980

WX  = sqrt 980   ≈ 31.3

It's correct

Here's the graph :

x^2 + (k - 9)x + 16

We need to have the discriminant =   0 

(k - 9)^2  - 4 * 16  = 0

(k -9)^2  - 64 =  0

(k -9)^2   = 64      take both roots

k - 9  =  8                    k -9  = -8

k  =17                        k = 1

Sum of k's   =  17 + 1  =  18

Let x be the total distance

Let B  = the distance he has driven

So

(1/2)x    = 12           multiply through by 2

x  = 24

2x^2 - 1x - 28  = -31 + jx

2x^2 - (1 + j )x + 3  = 0

We will have one root when  the discriminant =  0 

(1 + j)^2  - 4(2)(3)   =  0

j^2 + 2j + 1  - 24  =  0

j^2 + 2j - 23  = 0

j = [ -2 - sqrt [ 2^2 + 4*23 ] ] / 2

j =  [ -2 - sqrt [96] ] / 2

j = [ -2 - 4sqrt 6 ]  /2

j = - 1 - 2sqrt 6          and     j  =  -1 + 2sqrt 6

\(\frac{-21-\sqrt{201}}{10}\)

[-21 + sqrt (201) / 10   is also  a  root

product of the roots   = [  (-21)^2   - 201 ] / 100  =   240/100  =  2.4

product of roots =  v / 5

v / 5  = 2.4

v = 5 * 2.4    = 12

f(3)  = 4

So

(2/5)x + 3  = 3

x = 0

So

y  = 4f ( 3)  - 7   =

4 (4)  - 7   = 

9

(0, 9)  

is not one-to-one

[0, pi]