+0

0
162
3

1) Expressing your answer in interval notation, find all values of k such that the parabola y=x^2 + kx + 11 does not intersect the line y=2.

2) Expressing your answer in interval notation, find all values of s such that x(x-2)^2(x+1)<0.

3)Determine the range of the function g(x)=5x^2-2x+1. Enter your answer in interval notation.

Nov 2, 2019

#1
+2

1) Expressing your answer in interval notation, find all values of k such that the parabola y=x^2 + kx + 11 does not intersect the line y=2.

Let's see where the parabola DOES intersect y = 2

x^2 + kx + 11  = 2

x^2 + kx + 9  = 0

Where the discriminant is ≥ 0, we will have real solutions....so....

k^2 - 4*9  ≥ 0

k^2 ≥ 36

So.....the intervals  where the parabola will intersect the line y = 2  are when k =

(-infinity, -6 ] U [6, infinity )

So.....when    -6 < k < 6      the parabola WILL NOT intersect the line y = 2   Nov 2, 2019
#2
+2

2) Expressing your answer in interval notation, find all values of s such that x(x-2)^2(x+1)<0.

I think you must mean "x", not "s."

x ( x - 2)^2 (x + 1)  < 0

Note that when  x ≥ 0.... the result will be ≥  0

When  -1 < x < 0....the result will be < 0

When -inf < x < -1.....the result will be ≥ 0

So.....this will be < 0   on  the interval   ( -1, 0 )   Nov 2, 2019
#3
+2

3)Determine the range of the function g(x)= 5x^2-2x+1. Enter your answer in interval notation.

This is a parabola that turns upward

The x coordinate of the vertex  is    2/ [ 2 * 5]    =   1/5

The y coordinate of the vertex  is

5(1/5)^2 - 2(1/5) + 1   =

1/5 - 2/5 + 1  =

4/5

So...the range  is   [ 4/5, infinity )   Nov 2, 2019