# Given f(x) = -7x^2(2x-3)(x^2+1) how do you determine the following?

## a) The degree of $f \left(x\right)$. b) The leading coefficient. c) The maximum possible number of turning points. d) The real zeros of $f \left(x\right)$. e) The end behaviour of $f \left(x\right)$ as $x \to - \infty$ f) The end behaviour of $f \left(x\right)$ as $x \to \infty$

Aug 12, 2017

a) $\text{ } 5$
b) $\text{ } - 14$
c) $\text{ } 4$
d) $\text{ } 0$ of multiplicity $2$, $\frac{3}{2}$ of multiplicity $1$
e) $\text{ } f \left(x\right) \to \infty$ as $x \to - \infty$
f) $\text{ } f \left(x\right) \to - \infty$ as $x \to \infty$

#### Explanation:

Given:

$f \left(x\right) = - 7 {x}^{2} \left(2 x - 3\right) \left({x}^{2} + 1\right)$

It is not too arduous to multiply out $f \left(x\right)$ fully, but we do not need to in order to answer all of the questions.

In particular, note that the term of highest degree in $f \left(x\right)$ comes from multiplying:

$- 7 {x}^{2} \left(2 x\right) \left({x}^{2}\right) = - 14 {x}^{5}$

a) The degree of $f$ is the highest degree of any term, i.e. the degree of $- 14 {x}^{5}$, which is $5$. It is a quintic.

b) The leading coefficient is the coefficient of this term, namely $- 14$.

c) A polynomial of degree $n > 0$ has at most $n - 1$ turning points. So our quintic has at most $4$ turning points.

d) The real zeros of a polynomial correspond to its linear factors. In our example $f \left(x\right)$ has linear factors $x$, $x$ and $\left(2 x - 3\right)$, corresponding to real zeros $0$, $0$ and $\frac{3}{2}$. The quadratic factor $\left({x}^{2} + 1\right)$ will be non-zero for any real value of $x$, since ${x}^{2} \ge 0$ for any real value of $x$. The coincident zeros for $x = 0$ are described by saying that $f \left(x\right)$ has a zero at $x = 0$ of multiplicity $2$. The zero at $x = \frac{3}{2}$ has multiplicity $1$.

e) As $x \to - \infty$, the behaviour of $f$ is determined by the leading term $- 14 {x}^{5}$, which becomes much larger faster than the other terms. Since this term is of odd degree and the coefficient is negative ($- 14 < 0$), $f \left(x\right) \to \infty$ as $x \to - \infty$.

f) As $x \to \infty$, the behaviour of $f$ is dominated by the term of highest degree $- 14 {x}^{5}$, which becomes large and negative. So $f \left(x\right) \to - \infty$ as $x \to \infty$.

graph{-7x^2(2x-3)(x^2+1) [-5, 5, -25, 25]}