Question

Graphy y=x^ prime 2+3x+2 What are the zeros, the axis of symmetry, and the axordinates of the vertex? Show or explain your work for each

Answer 1

a) x=-2 or x=-1

b) x=-1.5

c) (-1.5,-0.25)

Explanation:

The given function is

`y = {x}^(2) + 3x + 2`

We complete the square as follows:

`y = {x}^(2) + 3x + 1.5 ^(2) - {1.5}^(2) + 2`

`y = {x}^(2) + 3x + 2.25 - 0.25`

We factor the perfect square;

`y = {(x + 1.5)}^(2) - 0.25`

We obtained the vertex form of the function.

a) To find the zeros, we set y=0.

`{(x + 1.5)}^(2) - 0.25 = 0`

`{(x + 1.5)}^(2) = 0.25`

`x + 1.5 = \pm √(0.25)`

`x = - 1.5 \pm 0.5`

`x = - 2 \: or \: x = - 1`

b) The parabola in the form

`y = a( {x - h)}^(2) + k`

has axis of symmetry at x=h.

Comparing this to

`y = ( {x + 1.5)}^(2) - 0.25`

We have

`x = - 1.5`

as the axis of symmetry.

The vertex is (h,k)=(-1.5,-0.25)