Question

3. The curve y= x^3-16x intersect x-axis at three points, then zeros of the curve are
(a) 0,0,0. b)-4,4,4
c) - 4,0,4. d)4,-4,4​

Answer 1

The correct option is C.

Step-by-step explanation:

The given equation is

$y=x^3-16x$

It intersect x-axis at three points.

The x-intercept are zeros of the function.

Equate y=0 to find the zeros of the function.

$x^3-16x=0$

$x(x^2-16)=0$

Using zero product property we get

$x=0$

$x^2-16=0\Rightarrow x^2=16\Rightarrow x=\pm4$

The zeros of the curve are -4, 0 and 4. Therefore, the correct option is C.

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Zeros of the polynomial

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Answer 2

Zero of the of the curve are 0,4,-4

So option (c) will be correct answer

Step-by-step explanation:

We have give the equation of the curve $y=x^3-16x$

When the curve intersect x axis then value of y = 0

So $x^3-16x=0$

Now taking x common $x(x^2-16)=0$

We know that $a^2-b^2=(a+b)(a-b)$

So $x(x+4)(x-4)=0$

x = 0,4,-4

So option (c) will be correct answer

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Find a point on the curve y=(x-2)² at which the tangent is parallel to the line joining the points (2,0) and (4,4)

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