anonymous
  • anonymous
Find x intercepts? y=3x^2+11x-4. Do I solve for y then substitute?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
set y=0 then solve
anonymous
  • anonymous
Y=0 for x intercepts.
anonymous
  • anonymous
let y = 0

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anonymous
  • anonymous
What does that mean?
anonymous
  • anonymous
\[\min{y = 3 x^2+11 x-4} = -169/12 at x = -11/6 \]
anonymous
  • anonymous
am i correct ?
anonymous
  • anonymous
3x(x+4)(-1(x+4)=3x-1=0=1/3 and x+4=0=-4 coming to: -1/3,-4 is this right?
anonymous
  • anonymous
or is this only y?
anonymous
  • anonymous
it also -11/6
anonymous
  • anonymous
and that the possible answer.
phi
  • phi
I assume you made a typo 3x-1=0=1/3 and x+4=0=-4 coming to: -1/3,-4 is this right? because the answer is +1/3 (which you found, but then typed in -1/3) @142343 x= -11/6 is where the parabola reaches a minimum, but its y-value at x= -11/6 is not zero. The question is asking for the 2 points where the parabola crosses the x-axis |dw:1327686228729:dw|
anonymous
  • anonymous
3x^2+11x-4=0 3x^2+12x-x-4=0 3x(x+4)-(x+4)=0 (x+4)(3x-1)=0 x=-4, 1/3

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