SOLUTION: Find the exact value for f(x)=3(x+4)^2-30. vertex= ,axis of symmetry= ,y-intercepts= ,x-intercepts= .

Question 200616: Find the exact value for f(x)=3(x+4)^2-30. vertex= ,axis of symmetry= ,y-intercepts= ,x-intercepts= .
Answer by RAY100(1637) (Show Source): You can put this solution on YOUR website!
f(x) = 3(x+4)^2 -30
y +30 = 3(x+4)^2
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std form,,,(y-k) = A ( x-h)^2
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-k=30,,,,k=-30,,,,,,,,,,-h = 4,,,h=-4,,,,Vertex ( h,k) = ( -4, -30)
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Axis of symmetry is ,,,,,,,, x=-4
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A is positive therefore open up (parabola)
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y intercept,,,,let x=0,,,,,y=3(0+4)^2 -30,,,,y= 48-30,,,,y=18
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x intercept,,,let y=0,,,,,0= 3(x+4)^2 -30 = 3 { (x+4)^2 -10}= x^2 +8x +16 -10 = x^2 +8x +6
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Use quadratic equation to solve,,,( note above 3 can be eliminated, due to 0 )
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a= 1,,,,b= 8,,,,,c=6
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x=[ -(8) +/- sqrt { (8)^2 - 4 (1)(6) } ] / 2(1)
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x =[ -8 +/- sqrt { 64 -24 } ] / 2
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x = [ -8 +/- sqrt (40) ] /2
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x= [ -8 +/- 6.32 ] /2
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x = -7.16,,,x= - .838,,,,,,,also solution to equation
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