Question

identify the interval(s) on which each quadratic function is positive. see example 5

1. y = x^2 + 9x + 18
2. y = x^2 + 2x - 8
3. y = x^2 - 5x - 24
4. y = -x^2 + 4x + 12
5. y = 2x^2 + 12x + 18
6. y = 5x^2 - 3x - 8

Explanation:

Step1: Find the roots of the quadratic function

For a quadratic function \(y = ax^{2}+bx + c\), we set \(y = 0\) and solve the quadratic - equation \(ax^{2}+bx + c=0\) using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) or by factoring.

Step2: Analyze the sign of the quadratic function in the intervals divided by the roots

If \(a>0\), the parabola opens upward, and the function is positive outside the interval of the roots. If \(a < 0\), the parabola opens downward, and the function is positive inside the interval of the roots.

For \(y=x^{2}+9x + 18\):

• Factor the quadratic equation: \(x^{2}+9x + 18=(x + 3)(x+6)=0\). The roots are \(x=-6\) and \(x=-3\). Since \(a = 1>0\), the function is positive for \(x<-6\) or \(x>-3\), i.e., \((-\infty,-6)\cup(-3,\infty)\).

For \(y=x^{2}+2x - 8\):

• Factor the quadratic equation: \(x^{2}+2x - 8=(x + 4)(x - 2)=0\). The roots are \(x=-4\) and \(x = 2\). Since \(a = 1>0\), the function is positive for \(x<-4\) or \(x>2\), i.e., \((-\infty,-4)\cup(2,\infty)\).

For \(y=x^{2}-5x - 24\):

• Factor the quadratic equation: \(x^{2}-5x - 24=(x-8)(x + 3)=0\). The roots are \(x=-3\) and \(x = 8\). Since \(a = 1>0\), the function is positive for \(x<-3\) or \(x>8\), i.e., \((-\infty,-3)\cup(8,\infty)\).

For \(y=-x^{2}+4x + 12\):

• First, factor out \(- 1\): \(y=-(x^{2}-4x - 12)\). Then factor \(x^{2}-4x - 12=(x - 6)(x+2)\). The roots of \(y = 0\) are \(x=-2\) and \(x = 6\). Since \(a=-1<0\), the function is positive for \(-2

For \(y = 2x^{2}+12x + 18\):

• Factor out \(2\) first: \(y = 2(x^{2}+6x + 9)=2(x + 3)^{2}\). The root is \(x=-3\). Since \(a = 2>0\), the function is positive for \(x

eq - 3\), i.e., \((-\infty,-3)\cup(-3,\infty)\).

For \(y=5x^{2}-3x - 8\):

• Factor the quadratic equation: \(5x^{2}-3x - 8=(5x - 8)(x + 1)=0\). The roots are \(x=-1\) and \(x=\frac{8}{5}\). Since \(a = 5>0\), the function is positive for \(x<-1\) or \(x>\frac{8}{5}\), i.e., \((-\infty,-1)\cup(\frac{8}{5},\infty)\).

Answer:

1. \((-\infty,-6)\cup(-3,\infty)\)
2. \((-\infty,-4)\cup(2,\infty)\)
3. \((-\infty,-3)\cup(8,\infty)\)
4. \((-2,6)\)
5. \((-\infty,-3)\cup(-3,\infty)\)
6. \((-\infty,-1)\cup(\frac{8}{5},\infty)\)