Question 1111168
The function's derivative, {{{"y '"}}} or {{{dy/dx}}} is
{{{dy/dx=-3x^2+4x+4}}} or {{{"y '"=-3x^2+4x+4}}} .
When {{{x=1}}} , {{{y}}} is indeed zero:
{{{y = - 1^3 + 2*1^2 + 4*1 - 5=-1+2+4-5=0}}} ,
so the graph does go through (1,0).
When {{{x=0}}} , the derivative's value is
{{{-3*1^2+4*1+4=-3+4+4=5}}} ,
so {{{5}}} is the slope of the tangent to the graph at (1,0) .
That means {{{y}}} is {{{highlight(increasing)}}} , and increasing steeply.
Obviously, if {{{y}}} is increasing around {{{x=1}}},
it is NOT decreasing.
 
The second derivative, {{{"y ' '"}}} or {{{d^2y/dx^2}}} ,
is the derivative of {{{"y '"=-3x^2+4x+4}}} ;
{{{"y ' '"=-6x+4}}} .
For {{{x=1}}} , the value of that second derivative is {{{-6*1+4=-6+4=-2}}} .
The fact that that value is negative means that {{{"y '"}}} is decreasing,
which means that the slope of the curve is increasing,
so it is growth rate is slowing,
and the curve is curling down.
It is {{{concave}}} {{{highlight(down)}}} , like a frown.
If the graph of {{{y=-x^3+2x^2+4x-5}}} is concave down,
it is obviously NOT concave up.
 
The {{{red(graph)}}} of {{{y=-x^3+2x^2+4x-5}}} ,
and the {{{green(tangent)}}} to that curve at (1,0) are shown below.
{{{graph(300,300,-4,6,-6.5,3.5,-x^3+2x^2+4x-5,5x-137/27)}}}
 
The graph above (and the one you could get in a graphing calculator),
shows that {{{y}}} is increasing at (1,0).
Because the tangent slope is so steep, {{{"y '> > 1"}} , 
and the curvature is so slight, {{{"y ' '"}}} not large,
it is not visually obvious that the curve is concave down,
although changing the scale and zooming helps:
{{{drawing(800,400,-0.1,1.5,-2.5,2.5,
green(line(0.5,-2.5,2,5)),
graph(800,400,-0.1,1.5,-2.5,2.5,-x^3+2x^2+4x-5)
)}}}