Find the vertex of the parabola whose equation in standard form is y = x2 + 8x - 1?

Question

help!?

Answer 1

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)

Answer 2

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)

Answer 3

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).

Answer 4

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).

Answer 5

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)

Answer 6

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)

Answer 7

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