Intercepts Touches or Crosses?

Question

For f(x) x^3 (x-1)^2 (x+6)^4, determine:

a. the x- and y- intercepts, if any

b. whether f(x) touches or crosses the x-axis at each intercept

My answer:

a. The intercepts are (0,0), (1,0), (-6,0). How do I find out which one crosses or touches. I need to be able to do this without using a graphical calculator.

Thanks.

Again, I would like to know how/why (1,0) and (-6,0) touches and (0,0) crosses. Thanks.

Answer 1

If you know calculus, take the derivative and check each point. If it has a slope of 0 then it will just touch. If it is positive or negative, it will cross.

Another (easier) way would be to just check x values to either side of the intercepts and calculate f(x) for them.. If they are both on one side, then it is a case of touching. If they are on opposite sides of the x-axis, we know that it must cross.

For example
x = -10 --> the result will be negative
x = -1 --> the result will be negative
x = 1/2 --> the result will be positive
x = 2 --> the result will be positive

So it will touch at (-6, 0), cross at (0, 0) and then touch at (1, 0).

Answer 2

a.
x-intercepts: x = 0, 1, -6
y-intercepts: y = 0

b.
f(x) touches at x = 1 and -6 and crosses at x = 0.
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Ideas: In factor x^3, you have an odd exponent. Therefore, it crosses x-axis at x = 0. In the other factors, you have even exponents. Therefore, f(x) touches x-axis at x = 1 and -6.