We want to know where the graph is above or below the x axis, in this case above because we are solving for 0. Plug in values to the equation and figure if they are above 0 or below 0: x x. 2x2 + 3 2 0 x. When you plug in and get 2 -s (like a value of -6), you get a + value, 1- and 1+ results in a negative value, and 2+s result in a + value. 5x2 + 3 3 2 + 2. Zeros: , -2 test=0= += test= 1=++=+ test= 3= 3+x 1 0 create a common denominator for the fraction and 1, then make it into one fraction. Solution: [0,3) (3 is not included because it is undefined at that point) zeros: 0 and 3 (undefined) 3 x 0 f (x ) = (x ) 1 / y2 then x in the domain of f where.
