Question

1. identify the interval(s) on which the function $y = x^2 - 2x - 48$ is positive. a $x < -6$ and $x > 8$ b $-6 < x < 8$ c $x > 6$ and $x < -8$ d $6 < x < 8$

Explanation:

Step1: Factor the quadratic function

To find where \( y = x^2 - 2x - 48 \) is positive, first factor the quadratic. We need two numbers that multiply to \(-48\) and add to \(-2\). Those numbers are \(-8\) and \(6\). So, \( y=(x - 8)(x + 6) \).

Step2: Find the roots

Set \( y = 0 \) to find the roots: \((x - 8)(x + 6)=0\) gives \( x = 8 \) or \( x=-6 \). These roots divide the number line into three intervals: \( (-\infty, -6) \), \( (-6, 8) \), and \( (8, \infty) \).

Step3: Test intervals

• For \( x < -6 \) (e.g., \( x=-7 \)): \( (-7 - 8)(-7 + 6)=(-15)(-1)=15>0 \), so the function is positive here.
• For \( -6 < x < 8 \) (e.g., \( x = 0 \)): \( (0 - 8)(0 + 6)=(-8)(6)=-48<0 \), so the function is negative here.
• For \( x > 8 \) (e.g., \( x = 9 \)): \( (9 - 8)(9 + 6)=(1)(15)=15>0 \), so the function is positive here.

Answer:

A. \( x < -6 \) and \( x > 8 \)