Question

What are the zeros, equation of the axis of symmetry and vertex of `y=x^2 + 5x+ 8`?

Answer 1

No real zeros.

Vertex `(-5/2, 7/4)` (minimum)

axis of symmetry `= -5/2`

FYI, this seems pretty hard for prealgebra.

Explanation:

To find the zeros (also called solutions or roots) if they exist, we set `y=0` and solve for x:

`y=x^2 + 5x+ 8`

`0=x^2 + 5x+ 8`

We have to use the quadratic formula to factor:

`ax^2 + bx +c`

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`a=1`
`b=5`
`c=8`

`x = (-5+-sqrt(5^2-4*1*8))/(2*1)`

`x = (-5+-sqrt(25-32))/(2)`

`x = (-5+-sqrt(-7))/(2)`

Since the `sqrt(-7)` is an imaginary number there all no real roots or zeros.

The formula for axis of symmetry (aos):

`aos=(-b)/(2a)`

`aos=(-5)/(2*1) = -5/2`

Fromula for vertex is:

`(aos, f(aos))`, remember `y=f(x)`

`f(x)=x^2 + 5x+ 8`

`f(-5/2)=(-5/2)^2 + 5(-5/2)+ 8`

`f(-5/2)=7/4`

So the vertex `= (-5/2, 7/4)`

Answer 2

The equation of the axis of symmetry is:

`x = -5/2`

The vertex is `(-5/2, 7/4)`

The zeros are:

`x_1 = -5/2-(sqrt7i)/2` and `x_2 = -5/2+(sqrt7i)/2`

Explanation:

When given a quadratic of the form `y = ax^2+bx+c`

The equation of the axis of symmetry is:

`x = -b/(2a)`

The x-coordinate of the vertex, `h`, has the same value as the axis of symmetry:

`h = -b/(2a)`

The y-coordinate of the vertex, `k`, is the function evaluated at `h`:

`k = ak^2+bk + c`

The zeros can be found using the quadratic formula:

`x_1 = (-b-sqrt(b^2-4(a)(c)))/(2a)` and `x_2 = (-b+sqrt(b^2-4(a)(c)))/(2a)`

Given: `y=x^2 + 5x+ 8`

Please observe that `a = 1, b = 1, and c = 8`

The equation of the axis of symmetry is:

`x = -5/2`

Compute the vertex:

`h = -5/2`

`k = (-5/2)^2+5(-5/2)+8`

`k = 7/4`

The vertex is `(-5/2, 7/4)`

Compute the zeros:

`x_1 = (-5-sqrt(5^2-4(1)(8)))/(2(1))` and `x_2 = (-5+sqrt(5^2-4(1)(8)))/(2(1))`

`x_1 = -5/2-(sqrt7i)/2` and `x_2 = -5/2+(sqrt7i)/2`