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# 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

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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)

Jun 8, 2019

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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   Jun 8, 2019
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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)   Jun 8, 2019
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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   Jun 8, 2019
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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   Jun 8, 2019