SOLUTION: Find the vertex and x-intercept(s) for the quadratic function. f(x)= -3x^2+2x+3 Round to the nearest hundredth if necessary. If there is more than one x-intercept, separate

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Find the vertex and x-intercept(s) for the quadratic function. f(x)= -3x^2+2x+3 Round to the nearest hundredth if necessary. If there is more than one x-intercept, separate      Log On


   

Question 395613: Find the vertex and x-intercept(s) for the quadratic function.
f(x)= -3x^2+2x+3
Round to the nearest hundredth if necessary.
If there is more than one x-intercept, separate them with commas.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

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

Finding x-intercepts when f(x)=0

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

x+=+%28-2+%2B-+sqrt%28+40+%29%29%2F%28-6%29+  |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%28x-h%29%5E2+%2Bk 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)