Question 1157414

 {{{f(x)=-3x^2+3x-2}}}

a. opens up or opens down

{{{opens}}}{{{ down}}}=> coefficient of the{{{ x^2}}} is negative

b. vertex =

write equation in vertex form

{{{f(x)=(-3x^2+3x)-2}}}

{{{f(x)=-3(x^2-x)-2}}}.......complete square

{{{f(x)=-3(x^2-x+b^2)-(-3)b^2-2}}}.......{{{b=1/2}}}

{{{f(x)=-3(x^2-x+(1/2)^2)+3(1/2)^2-2}}}

{{{f(x)=-3(x^2-x+(1/2)^2)+3(1/2)^2-2}}}

{{{f(x)=-3(x-1/2)^2+3/4)-2}}}

{{{f(x)=-3(x-1/2)^2+3/4-8/4}}}

{{{f(x)=-3(x-1/2)^2-5/4}}}

{{{h=1/2}}},{{{k=-5/4}}}

vertex is at ({{{1/2}}},{{{-5/4}}})


c axis of symmetry

The axis of symmetry always passes through the vertex of the parabola . The x -coordinate of the vertex is the equation of the axis of symmetry of the parabola.

{{{x=1/2}}}

d. show algebraically how you know the function has no x-intercepts

 {{{-3x^2+3x-2=0}}}

If the discriminant of a quadratic function is less than zero, that function has no real roots, and the parabola it represents does not intersect the x-axis.

if determinant <0
{{{b^2-4ac<0}}}
{{{3^2-4(-3)(-2)<0}}}
{{{9-24<0}}}
{{{-15<0}}}->proven



e. internal where it is increasing and decreasing
in interval ({{{-infinity}}},{{{1/2}}}) is increasing ,and in interval ({{{1/2}}},{{{infinity}}}) decreasing


{{{drawing ( 600, 600, -10, 10, -10, 10,
circle(1/2,-5/4,.12),locate(1/2,-5/4,v(1/2,-5/4)), green(line(1/2,10,1/2,-10)),
graph( 600, 600, -10, 10, -10, 10, -3x^2+3x-2)) }}}