Question

The points on the curve $y = 2 \times {x^2} - 6x - 4$ at which the tangent is parallel to the $x - axis$ is
$1)$ $\left( {\dfrac{3}{2},\dfrac{{13}}{2}} \right)$
$2)$ $\left( {\dfrac{{ - 5}}{2},\dfrac{{ - 17}}{2}} \right)$
$3)$ $\left( {\dfrac{3}{2},\dfrac{{17}}{2}} \right)$
$4)$ $\left( {0,-4} \right)$
$5)$ $\left( {\dfrac{3}{2},\dfrac{{ - 17}}{2}} \right)$

Answer 1

We have to find the points on the curve $y = 2 \times {x^2} - 6x - 4$ at which the tangent is parallel to the$\;x - axis$. We solve this question using the concept of derivatives and the concept of tangents of the curve . We first derivative $y$ with respect to x and then computing the derivative of $y$ to $0$ we find the values for $x$ . Then putting the value of $x$ in the given curve we find the value of y such that $\left( {x,y} \right)$is the point on the curve.

Complete step-by-step solution:
Differentiation, in mathematics , is the process of finding the derivative , or the rate of change of a given function. In contrast to the abstract nature of the theory behind it , the practical technique of differentiation can be carried out by purely algebraic manipulations , using three basic derivatives , four rules of operation , and a knowledge of how to manipulate functions. We can solve any of the problems using the rules of operations i.e. addition , subtraction , multiplication and division .
Given : $y = 2 \times {x^2} - 6x - 4$
For the tangent of the curve , we derive the curve with respect to.
Now we have to derivative of $y$ with respect to
Differentiating $y$ using the given rules of derivatives :
(Derivative of ${x^n} = n \times {x^{(n - 1)}}$)
(Derivative of constant$= 0$)
On differentiating , we get
$\dfrac{{dy}}{{dx}} = 4x - 6$
As , given in the question that the tangent of the curve is parallel to the $x-$axis this means that the slope of tangent of the curve and the $x - axis$ are equal .
The slope of $x - axis = 0$.
Using the relation , we get
$4x - 6 = 0$
From , this we get the value of $x$
So ,
$x = \dfrac{3}{2}$
Putting $x = \dfrac{3}{2}$in the curve , we get
$y = 2 \times {(\dfrac{3}{2})^2} - 6 \times (\dfrac{3}{2}) - 4$
$\Rightarrow y = 2 \times \left( {\dfrac{9}{4}} \right) - 6 \times \left( {\dfrac{3}{2}} \right) - 4\;$
On simplification,
$\Rightarrow y = \dfrac{9}{2} - 9 - 4$
This implies
$\Rightarrow y = \dfrac{{ - 17}}{2}$
Thus the point on the curve is $\left( {\dfrac{3}{2},\dfrac{{ - 17}}{2}} \right)$
Hence, the correct option is $\left( 5 \right)$.

Note: If $\dfrac{{dy}}{{dx}}$ does not exist at the point $\left( {{x_0},{y_0}} \right)$, then the tangent at this point is parallel to the y-axis and its equation is $x = {x_0}$. If tangent to a curve $y = f\left( x \right)$ at $x = {x_0}$ is parallel to x - axis , then $\dfrac{{dy}}{{dx}}$ at $\left( {x = {x_0}} \right) = 0$.