y=x^2-4x+20

Question

I have to find the window of the problem. Please help!

Answer 1

Are you asking what window settings to use to display this equation on a graphing calculator?

 

If so, look at the equation for information.

 

x^2 tells us the graph is a parabola.  Since the coefficient of x^2 is +1, we know that the parabola will open upward.

 

The constant (+20) tells us that the y-intercept is (0, 20). 

 

We can find the axis of symmetry by x = -b/2a, or x = -(-4)/2*1, or x = 2.  Substitute x = 2 into the equation to find the y-value of the vertex.  y = 2^2 - (4)(2) + 20 or y = 16.

 

Now we know the vertex is (2,16) and the vertex is a minimum.  So the graph will be limited to Quadrants I and II.  (Domain: x = all real numbers; Range: y >/= 16.) Using the standard window settings (-10, 10 scl 1 by (-10, 10 scl 1) will show none of this graph.  Changing the Y settings to reflect the Range of this graph will give you a good picture.

Answer 2

what do we mean by finding window of a problem here.please clarify.