Question

y = -x² - 1
which statement is true about the function?
○ when x is negative, y is positive
○ y is always 1 less than x
○ y > 0 for all values of x
○ y < 0 for all values of x
○ y = -3 when x = 1

Explanation:

Step1: Analyze the term \(x^2\)

For any real number \(x\), \(x^2\) is always non - negative, that is \(x^2\geq0\). When we multiply it by \(- 1\), we get \(-x^2\leq0\) (because multiplying a non - negative number by a negative number reverses the inequality).

Step2: Analyze the function \(y = -x^2-1\)

We know that \(-x^2\leq0\), then if we subtract 1 from both sides of the inequality \(-x^2\leq0\), we get \(y=-x^2 - 1\leq - 1\). Since \(y\leq - 1\), this means that for all real values of \(x\), \(y\lt0\).

Let's check the other options:

• Option 1: When \(x\) is negative, say \(x=-1\), then \(y=-(-1)^2 - 1=-1 - 1=-2\), which is negative. So this statement is false.
• Option 2: If \(x = 0\), \(y=-0^2-1=-1\), and \(x - 1=0 - 1=-1\), but if \(x = 1\), \(y=-1^2-1=-2\), and \(x - 1=0\), \(-2

eq0\). So this statement is false.

• Option 3: We have shown that \(y\leq - 1\), so \(y\) is never greater than 0. This statement is false.
• Option 5: When \(x = 1\), \(y=-1^2-1=-1 - 1=-2

eq - 3\). So this statement is false.

Answer:

\(y\lt0\) for all values of \(x\)