Question

Determine the equation of a cubic function with zeros at -2, -4, and 3 that passes through the point (4, 144)​

Answer 1

$\boxed{f(x)=3(x+2)(x+4)(x-3)}$

Explanation:

Since the cubic function has zeroes at:

$x=-2 \\ \\ x=-4 \\ \\ x=3$

Then, we can write this function as:

$f(x)=a(x-(-2))(x-(-4))(x-3) \\ \\ f(x)=a(x+2)(x+4)(x-3)$

So our goal is to find the leading coefficient $a$. Since that cubic function passes through the point (4, 144), then it is true that:

$f(4)=144 \\ \\ \\ So,\ plug \ these \values \ in \ the \ function: \\ \\ f(x)=a(x+2)(x+4)(x-3) \\ \\ 144=a(4+2)(4+4)(4-3) \\ \\ 144=a(6)(8)(1) \\ \\ 144=a(48) \\ \\ a=\frac{144}{48}=3$

Finally, the cubic function is:

$f(x)=3(x+2)(x+4)(x-3)$

The graph is shown below. As you can see the zeroes are:

$(-2,0) \\ \\ (-4,0) \\ \\ (3,0)$

And the function passes through $(4,144)$

Learn more:

Polynomial equations: brainly.com/question/1831722

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