CPhill

Using Vieta's Theorem

P + Q  =  -4       (1)

PQ  =  6           (2)

Square both sides of  (1)  and we have that

P^2 + 2PQ + Q^2 = 16          sub in (2)  for PQ   and we have

P^2 + 2(6) + Q^2  = 16

P^2 + Q^2  + 12  =  16

P^2 + Q^2  =  4

So b  = -4

And

(P^2)(Q^2)  = c

(PQ)^2  = c

(6)^2  =  c

So

b = - 4     and c =  36

And b + c  =  32

d. Solve f(x) > j(x)

-x^2 + 6x - 1  < 2x + 5      rearrange as

0 <   x^2 -4x + 6

x^2 - 4x + 6 > 0

This has no real solutions...so.....f(x)  is never > j(x)

The graph here confirms this :  https://www.desmos.com/calculator/v7jmn0b707

c. Reason out without calculation that the graphs of function f and j don't have any points in common

f = -x^2 + 6x - 1         j  = 2x + 5

Set these equal

2x + 5 = -x^2 + 6x - 1   rearrange as

x^2 -4x + 6  = 0

These cannot intersect  because  the discriminant   =  (-4)^2 - 4(6)  =  16 -24  = -8

And when the discriminant is < 0.....we have no real solutions

b. Show that the graphs of function h touches the graph of function f 

2x + 3  =  -x^2 + 6x - 1     rearrange as

x^2 - 4x + 4  = 0

(x + 2)^2  = 0

x = 2     and y = 2(2) + 3  = 7

So.....the function h touches f  at  (2, 7)

Given are the quadratic function f(x)= -x2+6x-1 and the three lineair functions g(x)=2x-1  h(x)=2x+3 and j(x)=2x+5

a. Calculate the coordinates of the points of intersection of the graphs of functions f and g

-x^2 + 6x - 1  = 2x - 1       rearrange as

0 =   x^2 -4x

x^2 - 4x  = 0        factor

x ( x - 4)  = 0

Set each factor to 0  and solve for x

x = 0         x - 4 = 0

                   x = 4

When x = 0,  y = 2(0) - 1  =  - 1       so  (0, -1)  is one intersection point

When x = 4, y = 2(4) - 1  = 7          so  (4, 7) is the other

a. For which values of x does g(x) > h(x) count?

g(x) > h(x)  on  (-inf, 1)  and on ( 4, inf)

b. For which values of x does f(x) > g(x) count?

f(x) > g(x)  on ( 0, 4)

h. Solve f(x) < h(x)

I think  h(x)  is  2x - 4

So....we want  to solve this

-(1/2)x^2 + 2x  <  2x - 4          subtract 2x from both sides

-(1/2)x^2 < - 4

0 < (1/2)x^2 - 4

(1/2)x^2 - 4 >  0          multiply through by 2

x^2 - 8 > 0 

x^2 > 8

This will be true when   (-inf, - sqrt (8) )   and  ( sqrt (8) , inf)

Correct  !!!

c. Calculate for which values of p the line has two points of intersection with the graph of g.

y = 2x + p      y  = x^2 - 2x  + 3

x^2 - 2x + 3  =  2x + p     rearrange   as

x^2 - 4x + (3 - p)  = 0           

If the discriminant is  >  0....we will have two intersection points so

(-4)^2  - 4(3 -p)  > 0

16 - 12 + 4p  > 0

4 + 4p > 0

1 + p > 0

p > -1

b. There is a line  that touches the graph of f. Calculate for which value of p the line touches the graph of f. 

y = -2x + p        and     y  = -2x^2 + 4x - 2      set these equal

-2x + p  = -2x^2 + 4x - 2     rearrange as

2x^2  -6x + (2 + p)  = 0

If the line just touches the quadratic.....then the discriminant must   =  0

So we have that

(-6)^2 - 4(2)(2 + p) = 0

36 - 8(2 + p)  = 0

36 - 16 - 8p  = 0

20 -8p  =0

20 = 8p       divide both sides by 8

20/8  = p   =  5/2

See the graph here :  https://www.desmos.com/calculator/cxek0jmo32

Given are the quadratic functions f(x) = -2x2 + 4x - 2 and g(x) = x2 - 2x + 3. Also is given the family of lines y = -2x + p

a. At p = 0,5 there is one of the lines. This line cuts the graph of f in two points. Calculate the x coordinates of those two points of intersections.

So we have 

y = -2x + .5        and     y  = -2x^2 + 4x - 2

Put the first function into the second for y   and we have that

-2x + .5  = -2x^2 + 4x - 2       rearrange as

2x^2 -6x + 2.5  = 0

2x^2 -6x + 5/2  = 0      multiply through by 2

4x^2 - 12x + 5  =  0   factor

(2x - 1) ( 2x - 5)  = 0

Set each factor to 0   and solve for x

2x -1  =0       2x - 5  = 0

2x = 1           2x  = 5

x = 1/2            x = 5/2 = 2.5