What are the intercepts of y = -x^2 + x ?

Answer 1

Y = -x^2 + x
y = x ( -x +1 )
if y = 0 => x ( -x +1 ) =0
x=0
OR
-x +1 = 0 = -x=-1 => x=1

the result is
x1=0
x2=1
y=0 and x1=0 (0,0)
y=0 and x2=1 (1,0)

Answer 2

The intercepts are x = 0, y= 0 and x = 1, y = 0

Answer 3

the intercepts of x

is when y = 0

-x^2 + x = 0

x (-x + 1) = 0


then x = 0 or

-x + 1 = 0

x = 1

intercepts of y is when x = 0

-(0)^2 + (0) = 0

then y intercept is (0,0)


intercepts = (0,0) and (1,0)

Answer 4

The intercepts are the value of x to make y=0, and the value of y when x=0. You should be able to solve.

Answer 5

There are two sets of intercepts to check for

y-intercepts (0,b)
set x = 0 and solve for "y"

y = -(0)^2 + 0
y = 0
y-intercept is (0,0)

x-intercepts (a,0)
set y = 0 and solve for "x"

y = 0 = -x^2 + x
0 = x(-x + 1)

x = 0
or -x + 1 = 0 -----> x = 1
x-intercepts are (0,0) and (1,0)

Intercepts: (0,0) and (1,0)

Answer 6

The intercepts are
(0,0) and (1,0)

Answer 7

y=-x^3 just guessing