For what values of is \(2x^2+8x\le-6\)? Express your answer in interval notation.
Add 6 to both sides: \(2x^2+8x+6\le 0\)
Divide by 2: \(x^2+4x+3\le0\)
Factorise: \((x+3)(x+1)\le 0\)
Critical values: \(x_1=-3,\text{and, }x_2=-1\)
Sketch the Quadratic, it is open-upwards and intersects the x-axis at x=-3 and x=-1.
Thus, the desired region is below the x-axis that is, between -3 and -1
Therefore, \(-3\le x\le-1\) is the desired solution.
In interval notation: \(x \in[-3,-1]\)
