Finding the vertex of a quadratic function?

Question

How do I find the vertex of and intercepts for this quadratic function?

 

y = x^2 + 2x - 3
 

Answer 1

To find the y-intercept, set x equal to 0. 0^2= 0 2*0 = 0 and you are left with y=-3

(0,-3) is the y-intercept.

Yo find the Vertex: x coordinate of vertex is -b/2a = -2 / 2 = -1

y coordinate of the vertex = (-1)^2+2(-1)-3=1-2-3 = -4

vertex is (-1,-4)

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You can use the quadratic equation with the following values to find the zeros:

a=1, b=2 and c=-3

Answer 2

The vertex of a quadratic equation is where x=-b/2a.

x= -2/2= -1

Vertex is @ x= -1.

Answer 3

How do I find the area of a missing piece of a shape?

Answer 4

As others have mentioned, the x value of the vertex is at the point where x = -b/2a

For quadratics with real roots, the x value of the vertex is the midpoint of the two roots (add the x values of the two roots and divide by 2).

You can then determine the y value of the vertex by plugging the x value back into the equation and 

Here's a worked solution with graph of y = x² + 2x - 3:

http://statsfiddle.info/Graph/SolveQuadraticEquation?QuadraticSolutionType=Coefficients&a=1&b=2&c=-3