Question

I’m not sure how to answer this in a Graph, Diagram, and Algebra(algebraic reasoning too) for a,b,c, and d. Can someone please help me?

Answer 1

The outline that that illustrate how k 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

$k > \dfrac{17}{4}$

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 y = 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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