you make x=0 and solve for y. for the x intercept you do y=0 and solve for x
x=-6+6=0 then x=-2+2=0 and 0+0=0
No. If you let 'x' =0 then 'x' is zero and there's no reason to solve for it.At the x-intercept ... just like at any point on the x-axis ... 'y' is zero.Let 'y' equal zero, then solve for 'x'.
Set it equal to 0 and subtract x and solve for y
factorise (x-5)(x+4)=0 Hence x= 5 or -4
you make x=0 and solve for y. for the x intercept you do y=0 and solve for x
16x³ - x = 0 By factorization, we obtain: x(16x² - 1) = 0 x(4x - 1)(4x + 1) = 0 Set each term by 0 and solve for x to get: x = 0 and 4x - 1 = 0 and 4x + 1 = 0 x = {0, ±¼}
10x=x 9x=0 x=0
substitute 0 for y and solve for x. then substitute x for 0 and solve for why and you have the x and y coordinates
0
9x2-9x = 0 x2-x = 0 x(x-1) = 0 x = 1 or x = 0
Suppose x3-4x = 0. To solve, factor: x3-4x = x(x2-4) = x(x+2)(x-2) = 0 Now, a product equals 0 if and only one or more of the factors equals 0, so set each factor to 0 and solve. The roots are 0,-2 and +2.
If that's x2 times 20 = 0, then x = 0, because only 0 x 20 = 0
x=-6+6=0 then x=-2+2=0 and 0+0=0
x2 + 7x = 0 => x*(x + 7) = 0 => x = 0 or x + 7 = 0 so that x = 0 or x = -7
x-intercept: 1. substitute 0 for y 2. solve for x y-intercept: 1. substitute 0 for x 2. solve for y
Set x = 0 and solve the resulting equation in y for the y-intercept. Set y = 0 and solve the resulting equation in x for the x-intercept.