CPhill

1)Suppose f is a quadratic function defined by f(x) = ax^2 + bx + c for some numbers a,b,and c. If g(x) = x - 2 and f(g(x)) = 2x^2 - 5x + 19 for all values of x, what is the value of a + b + c?

f(g(x)  must be    

a(x - 2)^2  + b(x - 2) +  c       expand this

a(x^2 - 4x + 4)  + bx - 2b  + c

ax^2  - 4ax  + 4a  + bx   - 2b  +  c

And we know that

ax^2  - 4ax  + 4a  + bx - 2b + c    =  2x^2  - 5x  +  19

ax^2  - ( 4a - b) x + (4a  - 2b + c)    =  2x^2  - 5x  + 19

It's obvious that  a   = 2

And   (4a - b)   = 5

So

4(2)  - b   = 5

8 - b  =  5

b   = 3

And

4a - 2b + c =  19

4(2) -2(3) + c  = 19

8 - 6  + c   =19

2 + c  =19

c  = 17

So

a + b  + c   =   2  +  3  +  17    =  22

URL  not found  ????

Note  that  if we let  g(x)  be  - 2x^5 + x^4  + ax^3

Then....adding  f(x) + g(x) we get

ax^3  + x^2  - 3    =    degree 3

So

g(x)  must also  be  degree 5

LOL!!!!!

Thanks, jfan  and CTG !!!!!

No....not 6

Look at the graph here :  https://www.desmos.com/calculator/b3vbllckme

Note that the roots of this polynomial  are  n = 3  and  n  = 6

So....at  n =6, the polynomial   = 0 

But we need it to  to be  negative......so.....the largest integer value to  make this true is  n   = 5

The slope of the first line  can be found as

3x + 4y  = -14

4y   =  -3x  - 14      divide  both sides by  4

y = (-3/4)x - 14/4

y = (-3/4)x - 7/2   

Since line 2  is perpendicular to this one then its slope =  4/3

And   since line 2 passes through  (-5, 7)   we have that

7 = (4/3)(-5) + b

7 = -20/3 + b

21/3  = -20/3  + b     add 20/3 to both sides

41/3  = b

So.... line 2  has the equation

y = (4/3)x + 41/3

And  m + b  =   (4/3) + (41/3)  =  45/3 =   15

We want to solve  this  :

n^2- 9n  +  18   <  0            factor the left side

(n  - 6) ( n - 3)  <  0

Note that this will  be true  when  n is on the interval  (3, 6)

So.....the greatest integer value of  n that makes the function negative is  n  = 5

The slope   of   Ax + By  =  3   is      -A/B    

And  x + 3y  = - 5  ⇒  3y  =  -x  - 5     divide through by 3  ⇒  y = (-1/3) - 5/3

So....the slope of the second line =  (-1/3)   

And since this is parallel to the first line,  this implies that

-1/3  = -A/B

So  A = 1   and  B    = 3

Then B - A  =   3  -  1   =   2

Let the measure of angle XAY   = x = angle BAC

Then, because AX = XY, then the measure  of angle XYA  also  = x  

And, by the exterior angle theorem , angle BXY  = 2x

And since XY =TB, then angle XBY  also equals 2x

So angle XYB =  180 -  angle BXY - angle XBY =   180 - 2x - 2x  = 180 - 4x

And  the sum  of angle XYA  +  angle XYB  + angle BYC = 180

So        x   + 180 - 4x + angle BYC  =180      (1)

So, rearranging (1).......  we have that    angle BYC  =   3x

And since YB = BC,  then the measure of angle BCY  also  =  3x = angle BCA

And in triangle BAC, the sum of angles  BAC  + angle ABC + angle BCA   =180

So   

x  +  120  +  3x    =180

4x  + 120  =  180

4x =  60

x = 15°  =   angle BAC