Question

Which of the following points is in the solution set of y < x2 - 4x + 3? (-3, 0) (1, 0) (2, -1)

Answer 1

The point (-3, 0) is in the solution set of y < x^2 - 4x + 3.

Step-by-step explanation:

We need to check each point to see if its y-coordinate is less than the corresponding x^2 - 4x + 3 value.

(-3, 0):

x^2 - 4x + 3 = (-3)^2 - 4 * (-3) + 3 = 9 + 12 + 3 = 24

Since 0 < 24, (-3, 0) is a solution.

(1, 0):

x^2 - 4x + 3 = 1^2 - 4 * 1 + 3 = 1 - 4 + 3 = 0

Since 0 < 0, (1, 0) is also a solution.

(2, -1):

x^2 - 4x + 3 = 2^2 - 4 * 2 + 3 = 4 - 8 + 3 = -1

Since -1 < -1, (2, -1) is not a solution.

Therefore, only (-3, 0) and (1, 0) are in the solution set of y < x^2 - 4x + 3.

Answer 2

(-3,0) and (2,-1)

Step-by-step explanation:

For (-3,0) x=-3 , y=0:

Putting in given inequality :

y < x2 - 4x + 3

0 < -3^2 -4*-3 +3 ("-" into "-" becomes "+")

0 < 9 +12 +3

0< 24 true

For (2, -1) x=2 , y= -1

Putting in given inequality :

y < x2 - 4x + 3

-1 < 2^2 -4(2) +3

-1 < 4 -6 +3

-1 < 7-6

-1 < 1 true

For (1,0) x=1 y=0

Putting in given inequality :

y < x2 - 4x + 3

0 < 1^2 -4(1)+ 3

0< 1- 4 + 3

0< 4 - 4

0 < 0 false

Hence, (-3,0) and (2,-1) are in the solution set of y < x2 - 4x + 3.

I hope it will help you!