Find the vertex point of each parabola?

Question

1) y = -(x+3)^2 - 2
2) y = -2x^2 +16x + 4
3) y = 3x^2 + x
4) x = y^2 - 4

An explanation would be appreciated
Thanks

Answer 1

A formula for the x-coordinate of the vertex is this

x = -b/2a

Then to find the y-coordinate of the vertex
substitute the x-coordiate that you get from the above equation into the original quadratic equation.

Example

y = 5x^2 + 10x - 7

vx is the x coordiate of the vertex

vx = (-b/2a)

vx = (-10/2(5)) = -1

vy is the y coordiate of the vertex

Now substitute into the quadratic equation to find the y-coordinate of the vertex.

y = 5(-1)^2 + 4(-1) - 7
= -6

vertex point = (vx,vy)

so the vertex point is (1,-6)

***Edit***
The vertex formula (x=-b/2a) I gave above works for all quadratics.
However there is a technique for quadratics that have the form of the first quadratic on you list.

If the quadratic is in the form

y = (x-h)^2 + c (where c is a constant)

The vertex of a quadratic in this form is always (h,c)

1.) set the (x-h)^2 equal to zero to find the x-coordinate of the vertex.

So (x-h)^2 = 0
x = h
Then you substitue the x-coordinate of the vertex back into the quadratic equation.

y = (-h+h)^2 + c

The y-coordinate of the vertex equals c.

So the vertex point is located at (h,c).

Answer 2

Find the vertex point of each parabola.

Use completing the square to find the vertex.

1) y = -(x + 3)^2 - 2
y + 2 = -(x + 3)^2

Vertex (h, k) = (-3, -2)

2) y = -2x^2 +16x + 4
y - 4 = -2(x^2 - 8x)
y - 4 - 2*16= -2(x^2 - 8x + 16)
y - 36 = -2(x - 4)²

Vertex (h, k) = (4, 36)

3) y = 3x^2 + x
y = 3(x^2 + x/3)
y + 3/36 = 3(x^2 + x/3 + 1/36)
y + 1/12 = 3(x + 1/6)²

Vertex (h, k) = (1/6, 1/12)

4) x = y^2 - 4
x + 4 = y^2

Vertex (h, k) = (-4, 0)

Answer 3

see http://www.analyzemath.com/quadratics/vertex_problems.html