Question

How do you find the minimum and maximum value of `y=(x+7)(x+3)`?

Answer 1

minimum `->(x,y)=(-5,-4)`
maximum `-> y=+oo`

Explanation:

If you multiply out the brackets you have the general form of:
`y=ax^2+bx+c` in this case `a=1` giving just `+x^2`

As this is positive the graph is of general shape `uu`
Thus the vertex is a minimum.

The vertex is `1/2` way between the x-intercepts.

Set:`" "y=0=(x+7)(x+3)`

Thus

`x+7 =0 => x=-7`
`x+3=0=>x=-3`

So `x_("vertex")=(-3-7)/2=-5`

Substitute `x=-5` to give:

`y_("vertex")=(-5+7)(-5+3)= 2xx(-2)=-4`

Thus the minimum `->(x,y)=(-5,-4)`

As the graph is of form `uu` then the maximum is `y=+oo`