# help me!! urgent!?

Find the coordinates of the vertex of each quadratic function.

solve and explain pls

1) y=x^2

vertex( , )

2)y=x^2+3

vertex( , )

3)y=3x^2

vertex( , )

4)y=3x^2-2

vertex( , )

5)y=-x^2+3

vertex( , )

6)y=x^2-6x+10

vertex( , )

7)y=3/4x^2-3x

vertex( , )

8)y=1/2x^2+2x+3

vertex( , )

### 2 Answers

- davidLv 71 month agoFavorite Answer
1. (0,0)

2. (0,3)

3. (0,0)

.... now you try the rest

I do not know what class you are taking .. there are a few different ways t do this.

Remember .. the general form is y = ax^2 + bx + c

.. the x value for the vertex is -b/(2a)

Once you know this value, sub. into the original equation to find the y value.

ex ... y = 3x^2 + 24x - 5

x val of vertex = -24 / (2*3) = -4

y val = 3(-4)^2 + 24(-4) - 5 = -53

vertex = (-4, -53)

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- Ian HLv 71 month ago
For parabolas a general method is more useful than individual answers.

If y = ax^2 + bx + c

dy/dx = 2ax + b = 0 for any vertex, (max or min)

Vertex is always at x = -b/2a

For example, y = x^2 - 6x + 10

x = -6/2 = -3, y = 9 + 18 + 10, Vertex (-3, 37)

y = 1/(ax^2 + bx + c) is not a parabola but it does have a peak.

set dy/dx = 0 and once again you find vertex is always at x = -b/2a

So, for example with y = 1/(2x^2 + 2x + 3), x = -2/4 = -1/2

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