Find the midpoint of the segment with the following endpoints. (9, 1) and (5, -7)
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The outline that that illustrate how <em>k</em> values change so that the solutions are the ones in the question are as follows
(a) The number line solution contains zero
Using algebra, we have;
The given quadratic inequality is; -x² + 4 ≥ x + k
The number line solution is the expression of the solution on the number line.
We have;
-x² + 4 ≥ x + k
-x² - x + 4 - k ≥ 0
Dividing by (-1), we get;
x² + x - 4 + k ≤ 0
When - 4 + k = -2
k = 2, we get;
x² + x - 4 + 2 ≤ 0
x² + x - 2 ≤ 0
(x - 1)·(x + 2) ≤ 0
Given that the result of the product is less than zero, we have;
((-3) - 1)·((-3) + 2) > 0
Therefore, the solution are;
x ≤ 1, or x ≥ -2
Which gives
-2 ≤ x ≤ 1
- Using diagram;
The solution of the above inequality includes 0 as shown in the attached number line
- Graphically;
The graph of the inequality is plotted using MS Excel, showing the solution set, of -2 ≤ x ≤ 1
(b) There are no solution to the inequality
In the present case, we have;
((-1)² - 4×(-1) × (4 - k) < 0
17 - 4·k < 0
Let k = 5, we have;
-x² + 4 ≥ x + k
-x² - x + 4 - 5 ≥ 0
-x² - x -1 ≥ 0
The discriminant, is (1² - 4×(-1)×(-1)) = -3 < 0, therefore, there are no real
roots to inequality, and -x² - x -1 is not equal to zero.
- Graphically
Please find attached the graph of the function, which does not intersect with the x-axis, and therefore, has no values at which <em>y</em> = 0
(c) The solution is a single number
-x² + 4 ≥ x + k
When k = x + 5, we have;
0 ≥ x² + 2·x + 1
0 ≥ (x + 1)²
Therefore;
The only solution is; x = -1
Using diagrams:
Please find attached the solution of the function on the number line
- Graphically
The graph of the function showing only one intercept is attached
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