Vertex form of a Quadratic Equation/Parabola

[Iceycold12]

Hello.

f(x) = a(x-h)^2 + k

Where (h,k) is the vertex.

H is negative by default in the equation, but that doesn't mean h can't be positive right? I believe having a negative H in the equation (x-(-h)) makes a positive h.

Thanks and excuse my horrid use of parenthesis, I believe the example is somewhat clear.

 

[wjm11]

Hello.

f(x) = a(x-h)^2 + k

Where (h,k) is the vertex.

H is negative by default in the equation, but that doesn't mean h can't be positive right? I believe having a negative H in the equation (x-(-h)) makes a positive h.

Thanks and excuse my horrid use of parenthesis, I believe the example is somewhat clear.

"h" is neither specifically positive nor negative in the equation; it is simply being subtracted from x. However, you are correct that h can be either positive or negative.

If you had y = (x - 3)^2, the "parent" function parabola (y = x^2), which has its vertex on the origin, would be slid three units to the right.

If you had y = (x + 3)^2, the parabola is slid three units to the left of the origin.

Hope that helps.

 

[Iceycold12]

Thanks wjm11.

 

[guitarguy]

The value k in your equation k shifts the graph vertically; up for positive k, down for negative k.

So the vertex would be at (h,k).