Want to ask us a question? Click here
Browse Questions
 Ad
0 votes

# The points at which the tangents to be curve $y=x^3-12x+18$ are parallel to x-axis are:

$\begin{array}{1 1} (A)\;(2,-2),(-2,-34) \\ (B)\;(2,34),(-2,0)\\(C)\;(0,34),(-2,0) \\ (D)\;(2,2),(-2,34) \end{array}$

Can you answer this question?

## 1 Answer

0 votes
Toolbox:
• Slope of a line is $ax+by+c=0$ is $- \bigg( \large\frac{ Coefficient\: of \: x}{ Coefficient\: of \: y} \bigg)$
• If $y = f(x),\: then \: \bigg( \large\frac{dy}{dx} \bigg)_p =$ slope of the normal to $y = f(x)$ at point $p$
Step 1
Equation of the curve is
$y = x^3-12x+18$
On differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}=3x^2-12$
If the tangent is parallel to $x$ - axis then
$\large\frac{dy}{dx}=0$
$\Rightarrow 3x^2-12=0$
$x^2=4$
$\Rightarrow x = \pm 2$
when $x = 2, \: y=2$
when $x = -2, \: y=-4$
Hence the points are $(2,2)\: and \: (-2, +34)$
The correct answer is D
answered Aug 9, 2013

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer