Functions- intervals and x intercepts

Question

f(x)=(x-1)(x-2)^2. Do I have 3 x-intercepts ...since if multiplied out the leading term would be x^3? Would I write 2- twice to get 3 intercepts.

How many intervals do I have when looking at x- intercepts? How would I write them?

Answer 1

Hi Cara

Setting your factors equal to zero will give you the x intercepts, there are 2. Also notice that when x is less than 1 y must be negative, when x is greater than 2 y must be positive and in between that is an interval where 1≤ x ≤2 where y is positive then goes back down to zero at x = 2, before going up again. You will see these if you graph your function

For

f(x) = (x -1 )(x - 2)²

Since (x - 2)² = (x - 2)(x - 2)

So your only factors are (x - 1) and (x - 2)

x - 1 = 0, x =1

x - 2 = 0, x = 2

x = 1 and x = 2 are you x intercepts

Edit, the root x = 1 has a multiplicity of 1 and root x = 2 has a multiplicity of 2

You should also calculate your y intercept which occurs when x is zero

y = (0 -1)(0 - 2)²

y = (-1)(-2)² = (-1)(4) = -4

You have three intervals

(-∞, 1)

(1, 2)

(2, ∞)

Graphing your function on a graphing calculator will confirm the intercept and you can see the intervals.

Answer 2

can be looked at as y=(x-2)(x-2)(x-1) so has three roots, x=2 has multiplicity of two

y=(x²-4x+4)(x-1)=x³-4x²+4x-x²+4x-4

=x³-5x²+8x-4

yes it is x³ but that means either y=0 at one x, or two x, or three x

in this case it is two x, x=1, and x=2

y=(x-1)³

can be looked at as y=(x-1)(x-1)(x-1) so x=1 is root with multiplicity of three

=x³-3x²+3x-1

is an example of x³ equation with only one x where y=0 in this case x=1

y=(x²+1)(x-1)=x³-x²+x-1

can be looked at as y=(x-i)(x+i)(x-1) so the three roots would be x=i, x=-i, and x=1

is another example of x³ equation with only one x on xy graph where y=0 in this case x=1