The standard form of a parabola is:
y = a(x - h)² + k
where (h, k) is the vertex. You've got
y = x² + 8x -1
So you need to complete the square:
y = x² + 8x + 16 - 16 - 1
y = (x + 4)² - 17
So the vertex is (-4, -17)
2006-12-28 06:30:36
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answer #1
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answered by Jim Burnell 6
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Find the vertex of the parabola whose equation in standard form is y = x2 + 8x - 1?
axis of symmetry is x= -b/2a = -8/2=-4
So y=16 -32 -1 = -17
So vertex is at (-4,-17)
2006-12-28 06:31:49
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answer #2
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answered by ironduke8159 7
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You need these 2 formulas: h= -b/2a and k=c-ah2. Plug in the numbers of the standard equation.
a=1
b=8
c=1
h=(-8)/(2(1))= -4
k=1-1(-4)2= 1-1(16)=-15
The vertex is (h,k), so the vertex for this equation is (-4,-15).
2006-12-28 06:34:59
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answer #3
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answered by Anonymous
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One way is to differiniate and set to 0.
So, you have 2x+8=0, or x=-4.
And y=-4^2+(8*-4)-1=16-32-1=-17.
So, vertex is at (-4,-17).
2006-12-28 06:31:42
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answer #4
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answered by yljacktt 5
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dy/dx = 2x + 8
2x + 8 = 0 at the vertex
x = -8/2 = -4
y = 16 - 32 - 1 from the original equation
y = -17
Vertex is (-4,-17)
2006-12-28 06:36:24
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answer #5
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answered by Anonymous
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y = x^2 + 8x - 1
y'=0=2x+8
2x=-8
x=-4
y=(-4)^2+8*(-4)-1=16-32-1=-17
vertex is at (-4, -17)
2006-12-28 07:05:05
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answer #6
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answered by yupchagee 7
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In y = ax^2 + bx + c, xv=-b/2a
In this case, xv= -8/2 = -4
To find yv, just substitute xv in the parabola equation
Once you have found xv and yv, you are done.
Ana
2006-12-28 06:28:20
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answer #7
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answered by MathTutor 6
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