Question 395613
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Hi
f(x)= -3x^2+2x+3
Finding x-intercepts when f(x)=0
 -3x^2+2x+3 = 0
{{{x = (-2 +- sqrt( 40 ))/(-6) }}}  |Using Calculator
x = -0.72, 1.39  |x-intercepts P(-.72,0) and Pt(1.39,0)
Finding vertex
Using the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
f(x)= -3x^2+2x+3   |completing square to put into vertex form
f(x) = -3[(x-1/3)^2 -1/9] + 3
f(x) = -3(x-1/3)^2 +1/3 + 3
f(x) = -3(x-1/3)^2 +10/3   Vertex is Pt(1/3,10/3)
{{{drawing(300,300, -6, 6, -6, 6, grid(1),
circle(-.72, 0,0.3),
circle(1.39, 0,0.3),
circle(1/3, 10/3,0.3),
graph( 300, 300, -6, 6, -6, 6, -3x^2+2x+3))}}}