y=(x-1)^2+4

Question

I need help with parabolas I am a teachers aid 

Answer 1

Common form of parabola learned in algebra is...

 

4p(y-k) = (x-h)²               For a vertical parabola

 

(h,k) is the vertex.

From the vertex, move a distance p upward to find the focus. (a point)

From the vertex, move a distance p downward to find the directrix. (a horizontal line)

 

The parabola is the curve equidistant from both focus and directrix.

 

y-4 = (x-1)²

(1,4) vertex, and p is 1/4

Answer 2

y=(x-1)^2+4 is the Vertex Form of a Quadratic Function that graphs as a parabola with Vertex (1,4), Axis of Symmetry x = 1, and is opening up.

 

You can find the Zeros of the function (if real they are the x-intercepts) by setting the function to zero and solving for x:

 

0 = (x-1)^2 + 4

 

-4 = (x-1)^2

 

√(-4) = 2i = |x-1|

 

x-1 = ±2i

 

x = 1±2i

 

In this case the Zeros are imaginary. There are no x-intercepts because the vertex is above the x-axis and opening up.

 

You can find the Standard Form of the Quadratic Function by expanding the Vertex Form:

 

y=(x-1)^2+4

y=x^2-2x+1+4

 

y=x^2-2x+5

 

You can find the Factored Form by using the Zeros:

 

y = (x-1+2i)(x-1-2i)