Question

How do you use end behavior, zeros, y intercepts to sketch the graph of `g(x)=(x+1)(x-2)^2(x-4)`?

Answer 1

graph{(x+1)(x-2)^2(x-4) [-10, 10, -5, 5]}

Explanation:

This is a quartic function. The equation has degree 4, so it is an even- degree polynomial.

The leading coefficient is positive, so as `x` approaches negative infinity, the graph rises to the left, and as `x` approaches positive infinity, the graph rises to the right.

In order to solve for the `y`-intercept, replace all of the `x` variables with 0's.

`y = (0+1)(0-2)^2(0-4)`

`y = (1)(4)(-4)`

`y = -16`

The constant term is -16, so the coordinate of the `y`-intercept is (0,-16).

The zeroes of the function are the `x`-intercepts of the graph. The `x`-intercepts of this equation are -1, 2, and 4.

The zero -1 has multiplicity 1, so the graph crosses the `x`-axis at the related `x`-intercept.

The zero 2 has multiplicity 2, so the graph just touches the `x`-axis at the related `x`-intercept.

The zero 4 has multiplicity 1, so the graph crosses the `x`-axis at the related `x`-intercept.