Question 1008646
<pre>
Instead of doing your problem for you, I'll do one exactly in every
detail step-by-step like yours.  All you have to do is use it as a
model and do the exact same steps.  So instead of your problem
I will do this one with only the numbers changed. I didn't even
change the 2 coefficient of x^2.

given {{{y=2x^2+bx+c}}}, the minimum value is -11 and the graph 
intersects at (0,-3). Find the values of b and c.

Since the equation passes through (0,-3) we substitute
(x,y) = (0,-3) into the given equation:

-3=2(0)²+b(0)+c
-3=0+0+c
-3=c

Therefore we already have the value of c, and

{{{y=2x^2+bx+c}}} becomes:

{{{y=2x^2+bx-3}}}

The minimum value of a quadratic equation is the y-coordinate
of the vertex. The formula for the x-coordinate of the vertex is

{{{-b/(2a)}}}

Since a=2 this becomes {{{-b/(2(2))=-b/4)}}}

Thus when we substitute the x coordinate of the vertex {{{-b/4}}},
we will get the y coordinate of the vertex which will be the
maximum value of -11.  So we substitute {{{-b/4}}} for x and -11
for y in

{{{y=2x^2+bx-3}}}
 
{{{-11=2(-b/4)^2+b(-b/4)-3}}}

Add 3 to both sides:

{{{-8=2(-b/4)^2+b(-b/4)}}}

{{{-8=2(b^2/16)-b^2/4)}}}

{{{-8=b^2/8-b^2/4)}}}

Multiply through by 8

{{{-64=b^2-2b^2}}}

{{{-64=-b^2}}}

{{{b^2=64}}}

{{{b= "" +- 8}}}

Thus there are two solutions for b, and 1 solution for c

Checking, we draw the graphs of

{{{y=2x^2+8x-3}}}  and  {{{y=2x^2-8x-3}}}

{{{drawing(350,400,-7,7,-13,3, locate(0,-2.5,"(0,-3)"),

locate(2,-11,"(2,-11)"),locate(-4,-11,"(-2,-11)"),

graph(350,400,-7,7,-13,3,2x^2-8x-3,2x^2+8x-3) )}}}

Edwin</pre>