Question

identify the interval(s) on which the quadratic function is positive.
y = 3x² - 22x + 24
select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
a. for all values of x in the interval < x <
b. for all values of x in the separate intervals x < and x >
c. for all values of x
d. for no values of x

Explanation:

Step1: Find the roots of the quadratic equation

To find where the quadratic function \( y = 3x^2 - 22x + 24 \) is positive, we first find its roots by solving \( 3x^2 - 22x + 24 = 0 \). Using the quadratic formula \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b=-22 \), and \( c = 24 \).

First, calculate the discriminant \( \Delta=b^2-4ac=(-22)^2-4\times3\times24=484 - 288 = 196 \).

Then, \( x=\frac{22\pm\sqrt{196}}{2\times3}=\frac{22\pm14}{6} \).

For the plus sign: \( x=\frac{22 + 14}{6}=\frac{36}{6}=6 \).

For the minus sign: \( x=\frac{22-14}{6}=\frac{8}{6}=\frac{4}{3} \).

Step2: Analyze the parabola's direction

The coefficient of \( x^2 \) is \( 3>0 \), so the parabola opens upwards. A parabola opening upwards is positive outside the interval between its roots. So the function is positive when \( x<\frac{4}{3} \) or \( x > 6 \).

Answer:

B. For all values of x in the separate intervals \( x<\frac{4}{3} \) and \( x>6 \)