Find the x-intercepts of the parabola with vertex (1,1) and y-intercept (0,-3). write your answer in this form (x of 1 , y of 1), (x of 2, y of 2). if necessary round to the nearest hundereth

Remember that the vertex form of a parabola or quadratic equation is:

y=a(x-h)^2+k, where (h,k) is the "vertex" which is the maximum or minimum point of the parabola  (and a is half the acceleration of the of the function, but that is maybe too much :P)

In this case we are given that the vertex is (1,1) so we have:

y=a(x-1)^2+1, and then we are told that there is a point (0,-3) so we can say:

-3=a(0-1)^2+1

-3=a+1

-4=a so our complete equation in vertex form is:

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

Now you wish to know where the x-intercepts are.  x-intercepts are when the graph touches the x-axis, ie, when y=0 so

0=-4(x-1)^2+1  add 4(x-1)^2 to both sides

4(x-1)^2=1  divide both sides by 4

(x-1)^2=1/4  take the square root of both sides

x-1=±√(1/4)  which is equal to

x-1=±1/2  add 1 to both sides

x=1±1/2

So x=0.5 and 1.5, thus the x-intercept points are:

(0.5, 0) and (1.5, 0)  or if you like fractions:

(1/2, 0) and (3/2, 0) :P

Sin(a)+cos(a)+cot(a)-1=2(csc(x)+1)