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
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)
+128475 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
+128475 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)
+128475 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
