Hi

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

Finding x-intercepts when f(x)=0

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

  |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,  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)



 

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