Question

19)The vertex of the parabola y=x^2+3x+2 lies in Quadrant
1. I
2. II
3. III
4. IV

Answer 1

Answer:

3.|||

Step-by-step explanation:

y=x²+3x+2

=x²+3x+9/4-9/4+2

=(x+3/2)²-(9-8)/4

=(x+3/2)²-1/4

put x+3/2=0

x=-3/2

then y=-1/4

so vertex is (-3/2,-1/4)

whiich lies in 3rd quadrant.

Answer 2

Answer:

option 3

Step-by-step explanation:

Given a parabola in standard form

y = ax² + bx +c ( a ≠ 0 ) , then

the x- coordinate of the vertex is

x = - $\frac{b}{2a}$

y = x² + 3x + 2 ← is in standard form

with a = 1, b = 3 , then

x = - $\frac{3}{2}$

Substitute this value into the equation for corresponding y- coordinate

y = (- $\frac{3}{2}$ )² + 3( - $\frac{3}{2}$ ) + 2 = $\frac{9}{4}$ - $\frac{9}{2}$ + 2 = - $\frac{1}{4}$

vertex = ( - $\frac{3}{2}$, - $\frac{1}{4}$ ) ← which is in the third quadrant