Question

How do you find the axis of symmetry, and the maximum or minimum value of the function `y = x^2 + 3x`?

Answer 1

Of form that has a minimum duo to the coefficient of `x^2` being positive.

`=>P_("minimum")->(x,y)=(-3/2,-9/4)`

Axis of symmetry is: `x=-3/2`

Explanation:

The `x^2` is positive so the general shape of the graph is `uu` thus we have a minimum.

Just for a moment, suppose we had `-x^2` in that scenario the graph would be of form `nn` and thus a maximum.

Consider the general case of `y=ax^2+bx+c`

Write this as: `a(x^2+b/ax)+c`

The axis of symmetry is at `x=(-1/2)xx(b/a)`

So in this case we have:

`x_("symmetry") = (-1/2)xx3/1 = -3/2 = x_("minimum")`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Set `x =-3/2" "=>" "y_("minimum")=(-3/2)^2+3(-3/2) = -2 1/4`

`=>P_("minimum")->(x,y)=(-3/2,-9/4)`

Answer 2

Axis of symmetry `x=-3/2`
Minimum value `-9/4`
Maximum value `oo`

Explanation:

To put the equation in standard form, x-terms are made into a perfect square. In the given expression consider the coefficient of x, which is 3. Half of this is `3/2`. Square this to have `9/4` Now add and subtract `9/4` on the right side of the expression, as shown below

#y= x^2 +3x +9/4 -9/4 =(x+3/2)^2 -9/4#

The equation in this form represents a vertical parabola, opening up, with vertex at `(-3/2, -9/4)`

The axis of symmetry is `x = -3/2`

Minimum value is `-9/4`, maximum value is `oo`