Anonymous

# vertex of parabolas?? HELP!!?

how do i find the vertex of the parabola represented by each function?

y = x^2

y = x^2 + 5

y = x^2 - 3

y = x^2 - 2

Update:

also for completing the square..i only know how to do it with ax^2 + bx + c. how would you do it with

x^2 + 4x

and

x^2 - 12x

can you explain the steps?? thanks!!!!

Relevance

the standard position is (0,0)

standard for of a parabola is y=x^2.

when you add or subtract things at the end it moves the vertex up and down so y=x^2 + 5, that vertex would be at (0,5)

For the next one it would be (0,-3) because the minus three brings the vertex down three points.

Adding and subtracting things at the end of the formula only moves the vertex up and down on the Y axis as much as is added or subtracted

• Anonymous
4 years ago

I can find the vertex of the parabola, but I am unaware of any curve named 'vertex parabola !' f(x) = x^2 - 4x + 3 = (x - 2)^2 - 1 The vertex is at: (2, - 1) QED

• A parabola of the form

y-k = (x-h)^2 or y=(x-h)^2+k has (h,k) as the vertex.

In all of your problems, the x-coordinate of the vertex is 0 as x^2=(x-0)^2

y=x^2 vertex(0,0)

y=x^2+5 vertex (0,5)

y=x^2-3 vertex (0,-3)

y=x^2-2 vertex (0,-2)

To complete the square

x^2+4x, divide the coefficient of x by 2.

(x+2)^2=x^2+4x+4

so, x^2+4x = (x+2)^2-4

(x^2-12x) same principle.

x^2-12x=(x-6)^2-36

• The general quadratic equation y = f(x) = ax^2 + bx + c represents a parabola.

To find the vertex, you have to find out the first derivative of f(x)

equal this to zero, and solve for x.

Therefore,

f'(x) = 2ax + b

2ax + b = 0

x = - b/2a

Now you plug this value into f(x) to calculate the other coordinate, and complete your vertex.

To complete the squares, you proceed always the same way...

• Anonymous

There is another way. -b/2a will give you the x-value for the vertex. In your problems, it looks like b=0 and a=1. Then substitute your answer for -b/2a for x and solve for y.

for x^2-12x. c=0. ignore c. b=12 and a=1 again.

• use -b

__

2a

to find the the x value of the vertex. Plug it into the equation to find the y-value

• Anonymous

gosh. um............. well isnt a vertex a point on a line? so if dere was a . on a ), den i guess dat would b it. gosh, y do ppl always ask calculus questions?

• ah pre calculus im in that right now

try this web site it will explain how