Question 670293

in general, the graph of a quadratic equation {{{y=ax^2+bx+c}}} is a parabola.

if a>0, then the parabola has a {{{minimum}}} point and it {{{opens}}} {{{upwards}}} (U-shaped)

the vertex: the x-coordinate of the minimum point (or maximum point) is given by

    {{{x=-b/2a}}}

then we substitute this x-value into our quadratic function (the {{{y}}} expression), solve it and we will have the ({{{x}}},{{{ y}}}) coordinates of the minimum (or maximum) point which is called the vertex of the parabola

so, the minimum point of the graph of the equation {{{y=2x^2+8x+9}}} will be:
 
the x-coordinate of the minimum point: {{{x=-b/2a}}}


{{{x=-8/2*2}}}

{{{x=-8/4}}}

{{{highlight(x=-2)}}}

the y-coordinate of the minimum point:

{{{y=2(-2)^2+8(-2)+9}}}

{{{y=2*4-16+9}}}

{{{y=8-16+9}}}

{{{highlight(y=1)}}}


{{{drawing(600,600,-10,10,-10,10,grid(1),locate(-2,1-.2,"V(-2,1)"),circle(-2,1,0.2),graph(600,600,-10,10,-10,10,2x^2+8x+9))}}}