# vertex of a curve

• May 8th 2013, 08:49 PM
Oldspice1212
vertex of a curve
Hey guys, been a while since I've done a question like this, a bit rusty was hoping you could help me clear some things.

Identify the type of curve and sketch the graph. I know it's a parabola, but having some trouble with it. I have this so far;

y = x^2+2x

0 = x^2 +2x

0 = x(x+2)

x = 0, and -2

So the y - int is (0,0)

If I want to sketch a graph though how would I go about doing it, and how do I get the vertex without using derivatives?

Thanks.
• May 8th 2013, 09:07 PM
ibdutt
Re: vertex of a curve
remember that quadratic equation y = ax^2+bx + c represents a parabola.
we can convert it to what is called a Vertex form y = a(x-h)^2 + k and then the point (h,k) is the vertex of the parabola and x= h is the axis of the parabola.
in this case we have
y = x^2 + 2x = x^2 + 2x + 1 -1= ( x+1)^2 -1
compare what is h and k and you have the coordinates of the vertex
• May 8th 2013, 09:11 PM
Oldspice1212
Re: vertex of a curve
Awesome thanks!
• May 8th 2013, 09:23 PM
MarkFL
Re: vertex of a curve
Another method is to observe that the vertex will lie along the axis of symmetry which in this case we can find simply by taking the vertical line midway between the roots, or $x=-1$. Hence, the vertex is at:

$(-1,y(-1))=?$