What are the coordinates of the vertx of the parabola y=x^2+4x-6?

Question

Y=x²+4x-6

Answer 1

If y = Ax² + Bx + C, A≠0, then the x-coordinate of the vertex is            -B/(2A)

 

So, for y = x²+4x-6, the x-coordinate of the vertex is -4/(2(1)) = -2

 

y-coordinate of vertex = (-2)²+4(-2)-6 = -10

 

vertex = (-2,-10)

 

 

Answer 2

The vertex form of a quadratic equation is y = a(x-h)² + k where (h,k) is the location of the vertex.  The equation in the problem is in standard form. To convert of to vertex form, you must "complete the square":

 

     y = x² + 4x - 6

     y = (x²+ 4x + (-4/2)² ) - 6 - (-4/2)2

     y = (x²+ 4x + 4) - 6 - 4

     y = (x+2)² - 10  which is now in vertex form.  The vertex is located at (-2, -10).

 

Another way to do it is to remember that when a quadratic is in standard form, y = ax2 + bx + c, the x-coordinate of the vertex is located at x = -b/2a.  In your equation, b = -4 and a = 1, so the x-coordinate of the vertex is located at x = -(4)/2*1 = -2.  To find the y-coordinate of the vertex, plug x = -2 into the equation.