theses dont make sense i dont get it help please?

Question

Find the coordinates of the vertex.

f(x) = 4x2 + 48x + 8

f(x) = x2 + 8x + 12


f(x) = x2 + 14x + 45

Answer 1

The x -coordinate of the vertex is -b/2a.
So for the first problem x = -48/(2*4) =- 6
so y = 4(-6)^2 +48(-6) + 8 = -136
Vertex is at (-6, -136)

2nd problem x coordinate = -8/2 = -4
So vertex is at (-4, f(-4))

3rd problem x= -14/2 = -7
So vertex is at (-7, f(-7)

Answer 2

if you use the standard form f(x)=ax^2+bx+c, you have the cordinates of the vertex to be ( -b/2a, f(-b/2a))

for example the first problem
4x^2+48x+8 the a is 4, the b is 48, the c is 8,
so (-b/2a)= (-48/2*4) =-48/8... so the x cordinate is -6, and plug it back in the the formula, 4(-6)^2+48(-6)+8 =-136

so (-6, -136) for the first one...

if you use the same way to solve the other two problems, you will have
x^2+8x+12----> (-4,-4)

and
x^2+14+45 -----> (-7,-4)

Answer 3

vertex is when f '(x) = 0

f(x) = 4x2 + 48x + 8
f'(x) = 8x + 48 = 0
or x = -6

Now f(1) = 4 * (-6)^2 + 48 * (-6) + 8
= 144 - 288 + 8
= -136

So the vertex is (-6, -136)

Answer 4

arrange the equation in standard form y - k = a(x-h)^2

where (h,k) is vertex

y = 4x^2 + 48x + 8

y=4(x^2+12) + 8

complete the square

y=4(x+6)^2 - 144 + 8

y = 4(x+6)^2 - 136

y + 136 = 4(x+6)^2

so vertex (-6,-136)

2)
y = x^2 + 8x + 12

y = (x+4)^2 - 16 + 12

y = (x+4)^2 - 4

y+4 = (x+4)^2

vertex(-4, -4)

3)

y = x^2 + 14x + 45

y = (x+7)^2 - 49 + 45

y = (x+7)^2 - 4

y + 4 = (x+7)^2

vertex = (-7, -4)