Can someone help me with this question? It seems simple but i keep getting the wrong answer

)The vertex of the parabola y=x^2+3x+2 lies in Quadrant

1. I

2. II

3. III

4. IV

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet

42.4

step-by-step explanation:

convert to decimal