Question 1040224
When the range of x determined by 
{{{-ax^2+bx+4>=0}}} is {{{-1/3<=x<=4}}}

then a = ?,  b = ?
<pre><b>In order for the function 

{{{y = -ax^2+bx+4}}}

to be greater than or equal to 0, its graph must be on or above the
x-axis.  A quadratic equation is the equation of a parabola.  

So this is a parabola that opens downward
and has x-intercepts (real zeros) at -1/3 and 4.

{{{graph(3400/21,400,-1.9,4.9,-1.9,14.9,(-3x^2+11x+4)*(sqrt(x+.347)/sqrt(x+.347))*(sqrt(4.018-x)/sqrt(4.018-x))) }}}



So we are to find the equation of the function

{{{y = -ax^2+bx+4}}}

In order for it to have zeros -1/3 and 4,
it must be of the form

{{{y = -a(x+1/3)(x-4)}}} 

{{{y = -a(x^2-4x+expr(1/3)x-4/3)}}}

{{{y = -ax^2+4ax-expr(1/3)ax+expr(4/3)a}}}

{{{y = -ax^2+(4a-expr(1/3)a)x+expr(4/3)a}}}

We compare it to

{{{y=-ax^2+bx+4}}}

The last (constant) terms must be the same, so

{{{expr(4/3)a=4}}}
{{{4a=12}}}
{{{a=3}}}

Substituting in

{{{y = -ax^2+(4a-expr(1/3)a)x+expr(4/3)a}}}

{{{y = -3x^2+(4*3-expr(1/3)*3)x+3*expr(4/3)3}}}

{{{y = -3x^2+(12-1)x+4}}}

{{{y = -3x^2+11x+4}}}

So a = 3 and b = 11

Edwin</pre></b>