Someone me with these 2 problems : (

If the discriminant b^2-4ac is 0, then you have TWO EQUAL, REAL ROOTS.

If you're given the x-intercepts, you can determine the factors of the polynomial as follows:   Take -3, change the sign and write (x+3).  Take 5, change the sign and write (x-5).  Then the eq'n of the parabola is

f(x) = (x+3)(x-5) = x^2 - 2x -15, in which a=1, b = -2 and c= -15.

You can find the x-coordinate of the vertex, which is also the equation of the axis of symmetry, using

x= -b / (2a).    Here, x = -(-2) / (2[1]), or  x = 1

Find the y-coordinate by subbing 1 for x in the equation above:

y = (1)^2 - 2(1) - 15  =  1 - 2 - 15  =  -16

The vertex is at (1, -16) and the equation of the axis of symm. is x = 1.

You need to show the box plots

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