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

Hint: 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 \[\left( {x = {x_0}} \right) = 0\].