Question

How do you find the stationary points of the function $y={{x}^{2}}+6x+1$ ?

Answer 1

To find the stationary points of the function $y={{x}^{2}}+6x+1$ , we have found its derivative and equate it to zero. This will yield to the equation $2x+6=0$ . Then we have to solve for x. We will then substitute this in the given function to find the value of y. Then the stationary point will be $\left( x,y \right)$ .

Complete step-by-step solution:
We need to find the stationary points of the function $y={{x}^{2}}+6x+1$ . First, let us see what stationary point is. Stationary point of a function is a point on the graph of the function where its derivative is equal to zero. This means that if f(x) is a function, then $\dfrac{d}{dx}f(x)=0$ or $f'\left( x \right)=0$ .
Now, let us consider the function $y={{x}^{2}}+6x+1$ . We have to take the first derivative of this function.
$\dfrac{dy}{dx}=2x+6$
We know that the stationary points of y will be when $\dfrac{dy}{dx}=0$ . Therefore, we can write
$\dfrac{dy}{dx}=2x+6=0$
Now, we have to solve for x. Let us take the constant term to RHS.
$2x=-6$
Let’s find x by taking 2 to the RHS. We will get
$x=\dfrac{-6}{2}=-3$
Now, we have to find the y coordinate. For this, we have to substitute the value of x in the given function. We will get
$\begin{aligned} & y={{x}^{2}}+6x+1 \\\ & \Rightarrow y={{\left( -3 \right)}^{2}}+6\times \left( -3 \right)+1 \\\ \end{aligned}$
On solving, we will get
$y=9-18+1=-8$
Hence, the stationary point occurs at $\left( -3,-8 \right)$.

Note: Students must be thorough with the derivatives and rules of derivatives. Students have a chance of making mistakes by stopping the solution after finding x. Never forget that a point contains two variables, that is, $\left( x,y \right)$ .