# 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( , )

Relevance

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)

• Login to reply the answers
• 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

• Login to reply the answers