Given are the quadratic function f(x)=-x^{2}+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

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

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

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

Guest Jun 8, 2019

#1**+2 **

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

CPhill Jun 8, 2019

#2**+2 **

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)

CPhill Jun 8, 2019

#3**+1 **

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

CPhill Jun 8, 2019