1. Determine which ordered pair is a solution of y = x^2-2
(5, 21),
(–5, 24),
(–1, –1), ••
(–2, 6)
2. Without graphing, tell whether the line containing (6, –3) and (0, –8) is horizontal, vertical, or neither
I think it's horizontal
Correct my answers
#1, correct
#2, to be horizontal, they must have the same y
to be vertical, they must have the same x
So, do you wish to change your answer?
To determine whether a given ordered pair is a solution to an equation, you need to substitute the values of x and y into the equation and see if it satisfies the equation. Let's apply this process to question 1:
1. Determine which ordered pair is a solution of y = x^2-2:
- For (5, 21), we substitute x = 5 and y = 21 into the equation:
y = 5^2 - 2 = 25 - 2 = 23.
Since 23 does not equal 21, (5, 21) is not a solution.
- For (–5, 24), we substitute x = -5 and y = 24 into the equation:
y = (-5)^2 - 2 = 25 - 2 = 23.
Since 23 does not equal 24, (–5, 24) is not a solution.
- For (–1, –1), we substitute x = -1 and y = -1 into the equation:
y = (-1)^2 - 2 = 1 - 2 = -1.
Since -1 equals -1, (–1, –1) is a solution.
- For (–2, 6), we substitute x = -2 and y = 6 into the equation:
y = (-2)^2 - 2 = 4 - 2 = 2.
Since 2 does not equal 6, (–2, 6) is not a solution.
Therefore, the ordered pair (–1, –1) is the solution to the equation y = x^2-2.
Moving on to question 2, to determine the nature of a line without graphing, we can examine the slopes of the line segments formed by the given points.
2. Without graphing, tell whether the line containing (6, –3) and (0, –8) is horizontal, vertical, or neither:
- We can calculate the slope of the line using the formula: slope = (change in y) / (change in x).
- For the given points (6, –3) and (0, –8), the change in y is: (-8) - (-3) = -8 + 3 = -5.
- The change in x is: 0 - 6 = -6.
- Therefore, the slope of the line is: slope = -5 / -6 = 5/6.
- Since the slope is not zero, the line is neither horizontal nor vertical.
Based on this analysis, the line containing (6, –3) and (0, –8) is neither horizontal nor vertical.