Question 1192428
<br>
{{{f(x)=2x^3+ax^2+b}}}<br>
{{{f(-3)=19}}}
{{{-54+9a+b=19}}}<br>
{{{df/dx=6x^2+2ax}}}<br>
The stationary point at x=-3 is where the derivative is zero.<br>
{{{6x^2+2ax=54-6a=0}}}
{{{a=9}}}<br>
Use that value of a to solve for b.<br>
{{{-54+9(9)+b=19}}}
{{{27+b=19}}}
{{{b=-8}}}<br>
The function is {{{2x^3+9x^2-8}}}<br>
We know f(-3)=19; and f(0)=-8.  Those two points, along with the positive leading coefficient, tell us that the stationary point at (-3,19) is a local maximum.<br>
Or, more formally, to show that the stationary point is a local maximum, we could find the second derivative of the function and show that it is negative at x=-3.<br>
ANSWERS:
a=9
b=-8
(-3,19) is a local maximum<br>
A graph....<br>
{{{graph(400,400,-5,3,-10,30,2x^3+9x^2-8)}}}<br>