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info@snapandsolve.comTo find the intervals where \( y = x^2 - 2x - 48 \) is positive, we first find the roots of the equation by setting it to zero:
\( x^2 - 2x - 48 = 0 \)
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -2 \), and \( c = -48 \):
\( x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-48)}}{2 \cdot 1} \)
\( x = \frac{2 \pm \sqrt{4 + 192}}{2} \)
\( x = \frac{2 \pm \sqrt{196}}{2} \)
\( x = \frac{2 \pm 14}{2} \)
\( x = 8 \) or \( x = -6 \)
The roots are \( x = -6 \) and \( x = 8 \). The parabola opens upwards (since the coefficient of \( x^2 \) is positive), so it is positive outside the interval between the roots.
Therefore, the intervals where \( y \) is positive are:
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