Determine the values of $x$ for which the function $f(x)=x^{2}-6 x+9$ is increasing or decreasing. Also, find the coordinates of the point on the curve $y=x^{2}-6 x+9$ where the normal is parallel to the line $y=x+5$.

Home » Determine the values of for which the function is increasing or decreasing. Also, find the coordinates of the point on the curve where the normal is parallel to the line .

Determine the values of $x$ for which the function $f(x)=x^{2}-6 x+9$ is increasing or decreasing. Also, find the coordinates of the point on the curve $y=x^{2}-6 x+9$ where the normal is parallel to the line $y=x+5$ .

Determine the values of $x$ for which the function $f(x)=x^{2}-6 x+9$ is increasing or decreasing. Also, find the coordinates of the point on the curve $y=x^{2}-6 x+9$ where the normal is parallel to the line $y=x+5$ .

Solution:

Given that $f(x)=x^{2}-6 x+9$
$\begin{array}{l} \Rightarrow \\ f(x)=\frac{d}{d x}\left(x^{2}-6 x+9\right) \\ \Rightarrow f^{\prime}(x)=2 x-6 \\ \Rightarrow f^{\prime}(x)=2(x-3) \end{array}$
For $f(x)$ let us find critical point, we must have
$\begin{array}{l} \Rightarrow f^{\prime}(x)=0 \\ \Rightarrow 2(x-3)=0 \\ \Rightarrow(x-3)=0 \\ \Rightarrow x=3 \end{array}$
So, $f^{\prime}(x)>0$ if $x>3$ and $f^{\prime}(x)<0$ if $x<3$
Therefore, $f(x)$ increases on $(3, \infty)$ and $f(x)$ is decreasing on interval $x \in(-\infty, 3)$
Let’s find the coordinates of point
Eq. of curve is $f(x)=x^{2}-6 x+9$
Slope of this curve is given by
$\begin{array}{l} \Rightarrow m_{1}=\frac{d y}{d x} \\ \Rightarrow m_{1}=\frac{d}{d x}\left(x^{2}-6 x+9\right) \\ \Rightarrow m_{1}=2 x-6 \end{array}$
Eq. of line is $y=x+5$
Slope of this curve is given by
$\begin{array}{l} m_{2}=\frac{d y}{d x} \\ \Rightarrow m_{2}=\frac{d}{d x}(x+5) \\ \Rightarrow m_{2}=1 \end{array}$
As slope of curve is parallel to line
So, they follow the relation
Therefore putting the value of $\mathrm{x}$ in equation of curve, we obtain
$\begin{array}{l} \Rightarrow y=x^{2}-6 x+9 \\ \Rightarrow y=\left(\frac{5}{2}\right)^{2}-6\left(\frac{5}{2}\right)+9 \\ \Rightarrow y=\frac{25}{4}-15+9 \\ \Rightarrow y=\frac{25}{4}-6 \\ \Rightarrow y=\frac{1}{4} \end{array}$
Therefore the required coordinates is $\left(\frac{5}{2}, \frac{1}{4}\right)$