Question 350830
Convert to vertex form, {{{y=a(x-h)^2+k}}}.
The max or min occurs at the vertex, min if {{{a>0}}}, max if {{{a<0}}}.
Using your example,
{{{y=(x+5)(x-3)}}}
Use the FOIL method to expand,
{{{y=x^2-3x+5x-15}}}
{{{y=x^2+2x-15}}}
Complete the square to get into vertex form,
{{{y=(x^2+2x+1)-15-1}}}
{{{highlight(y=(x+1)^2-16)}}}
So the vertex occurs at ({{{-1}}},{{{-16}}}).
Since {{{a=1}}}, the parabola opens upward and the vertex value is a minimum.
{{{ymin=-16}}}
.
.
.
{{{drawing(300,300,-5,5,-20,5,circle(-1,-16,0.2),grid(1),graph(300,300,-5,5,-20,5,(x+5)(x-3),(x+1)^2-16
))}}}