QuestionGiven the function $g(n)=3 n^{4}+12 n^{3}-63 n^{2}$ : its $g$ -intercept is $\square$ its $n$ -intercepts are $\square$

Studdy Solution

STEP 1

1. The function given is $g(n) = 3n^4 + 12n^3 - 63n^2$ .
2. The $g$ -intercept is the value of the function when $n = 0$ .
3. The $n$ -intercepts are the values of $n$ for which $g(n) = 0$ .

STEP 2

1. Find the $g$ -intercept.
2. Find the $n$ -intercepts by solving $g(n) = 0$ .

STEP 3

To find the $g$ -intercept, substitute $n = 0$ into the function:
$g(0) = 3(0)^4 + 12(0)^3 - 63(0)^2$
Simplify the expression:
$g(0) = 0$
Thus, the $g$ -intercept is $\boxed{0}$ .

STEP 4

To find the $n$ -intercepts, solve the equation $g(n) = 0$ :
$3n^4 + 12n^3 - 63n^2 = 0$
Factor out the greatest common factor, which is $3n^2$ :
$3n^2(n^2 + 4n - 21) = 0$
Set each factor equal to zero:
1. $3n^2 = 0$
2. $n^2 + 4n - 21 = 0$
Solve the first equation:
$3n^2 = 0$ $n^2 = 0$ $n = 0$
Solve the second equation using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 1$ , $b = 4$ , $c = -21$ :
$n = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-21)}}{2 \cdot 1}$ $n = \frac{-4 \pm \sqrt{16 + 84}}{2}$ $n = \frac{-4 \pm \sqrt{100}}{2}$ $n = \frac{-4 \pm 10}{2}$
Calculate the two possible solutions:
$n = \frac{-4 + 10}{2} = \frac{6}{2} = 3$ $n = \frac{-4 - 10}{2} = \frac{-14}{2} = -7$
Thus, the $n$ -intercepts are $\boxed{0, 3, -7}$ .
The $g$ -intercept is $\boxed{0}$ and the $n$ -intercepts are $\boxed{0, 3, -7}$ .

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