A few days ago
Anonymous

# hello i need some help!?

I need help with these problem.. 1 or the other or both.. im stuck on them thanks for the help!!

Find the coordinates of the vertex for the parabola defined by the given quadratic function.

f(x) = -x2 – 8x – 4

Find the coordinates of the vertex for the parabola defined by the given quadratic function.

f(x) = (x + 2)2 – 7

A few days ago

to find the x value of the vertex of a quadric function. Use -b/(2a)

f(x) = -x^2 – 8x – 4

a = -1

b = -8

-(-8) / (2 * -1)

8/-2

-4

plug -4 back in the eqution to find the y value of the vertex

f(-4) = -(-4)^2 – 8(-4) – 4

= -16 + 32 – 4

= 16 – 4

= 12

so the vertext is (-4,12)

Vertex form:

f(x) = a(x – h) + k has the vertex of (h,k)

f(x) = (x + 2)^2 – 7 has the vertex of (-2,-7)

0

A few days ago
gabyrig
First:

f(x) = -x^2 – 8x – 4

a = -1

b = -8

c = -4

Second:

f(x) = (x+2)^2 – 7

Multiply out:

f(x) = x^2 + 2*2*x + 2^2 – 7

Simplify:

f(x) = x^2 +4x + 4 -7 = x^2 + 4x -3

Find your a, b and c (the sign in front always goes with the number) then plug into the formula the person above gave you and voila!

Otherwise if you know calculus, you know that the vertex is the very top where the curve goes from one curving direction to the other. The peak. The slope at that point is always zero so you can find the first derivative of each equation, set it to zero and that will give you the x (then plug into the formula again to get the y coordinate.

First:

f'(x) = – 2x -8 = 0

Solve for x

Second:

f'(x) = 2x + 4 =0

solve for x.

It’s faster but maybe not at your level.

0

A few days ago
elipra91
If you have any equation of a parabola it will look like this:

y=ax^2+bx+c

To find the vertex (x,y) do the following:

1- First find the x point by using this formula: -b/2a

2 – Use the result and substitute it by the x in the given formula and then you’ll have y.

3 – Done!!

First problem:

f(x)= -x^2-8x-4

a= -1 b= -8 c= -4

-b/2a = -(-8)/ 2 .(-1) = 8/-2= -4

x= -4

Substitute:

y= -(-4)^2-8.(-4)-4

y= -16 +32 – 4

y= -16 + 28

y= 12

Second problem:

Expand f(x)= (x+2)^2 – 7

y= x^2+4x-3

a= 1 b= 4 c= -3

-b/2a= -4/2 = -2

x= -2

Substitute:

y= 4 + 8 – 3

y= -7