How do you find a unit vector a) parallel to and b) normal to the graph of \[f(x)=-{{x}^{2}}+5\] at a given point (3,9)?

Hint: In this question, we will find the equation of slope and slope of tangent. After that, we will find the unit vector that is parallel and normal to the graph \[f(x)=-{{x}^{2}}+5\] using that slope. For solving this question, we should know that multiplication of slope of two perpendicular lines is always -1.

Complete step by step answer:
As we know that, the slope of any line \[f(x)\] is \[\dfrac{d\left[ f(x) \right]}{dx}\].
Hence, slope of \[f(x)=-{{x}^{2}}+5\] can be found by differentiating \[f(x)\].
Therefore,
\[\dfrac{d\left[ f(x) \right]}{dx}=\dfrac{d\left[ -{{x}^{2}}+5 \right]}{dx}=\dfrac{d\left[ -{{x}^{2}} \right]}{dx}+\dfrac{d\left[ 5 \right]}{dx}=-2x+0=-2x\]
Hence, the slope of the graph is -2x.
Part(a)
Now, we will check the slope of the point (3,9).
The slope of graph at (3,9) where x=3 and y=9 is
\[2x=-2\times 3=-6\]
Now, let us take a unit vector with slope = -6 and a base at origin.
\[\dfrac{y}{x}=-6\]
\[\Rightarrow y=-6x\]
As we have taken the unit vector. Due to property of a unit vector, we can say that
\[\sqrt{{{x}^{2}}+{{y}^{2}}}=1\]
Putting the value of y=-6x, we get
\[\Rightarrow \sqrt{{{x}^{2}}+{{\left( -6x \right)}^{2}}}=1\]
\[\Rightarrow \sqrt{{{x}^{2}}+36{{\left( x \right)}^{2}}}=1\]
\[\Rightarrow \sqrt{37{{x}^{2}}}=1\]
Squaring both sides, we have
\[\Rightarrow \sqrt{37}x=1\]
\[\Rightarrow x=\dfrac{1}{\sqrt{37}}\]
And \[\Rightarrow y=-\dfrac{6}{\sqrt{37}}\]
Unit vector that is parallel to graph is \[\overrightarrow{\left( \dfrac{1}{\sqrt{37}},-\dfrac{6}{\sqrt{37}} \right)}\] at (3,9).
Part (b)
Now, we will find the unit vector normal to the given graph.
We know that if two lines are perpendicular to each other, then the multiplication of slope of both of the lines is always -1. And normal and parallel lines of a graph are always perpendicular to each other.
Therefore, the slope of the normal which will be used to find the unit vector is \[\dfrac{1}{6}\].
Hence, the equation of normal is:
\[\dfrac{y}{x}=\dfrac{1}{6}\]
\[\Rightarrow y=\dfrac{x}{6}\]
Due to the property of the unit vector, we can say that its magnitude will be 1.
Therefore,
\[\sqrt{{{x}^{2}}+{{y}^{2}}}=1\]
Putting the value of \[y=\dfrac{x}{6}\], we get
\[\Rightarrow \sqrt{{{x}^{2}}+{{\left( \dfrac{x}{6} \right)}^{2}}}=1\]
\[\Rightarrow \sqrt{{{x}^{2}}+{{\dfrac{x}{36}}^{2}}}=1\]
\[\Rightarrow \sqrt{\dfrac{37{{x}^{2}}}{36}}=1\]
\[\Rightarrow \dfrac{\sqrt{37}x}{6}=1\]
\[\Rightarrow x=\dfrac{6}{\sqrt{37}}\]
Then, putting the value of x in \[y=\dfrac{x}{6}\], we get
\[\Rightarrow y=\dfrac{6}{\sqrt{37}}\times \dfrac{1}{6}=\dfrac{1}{\sqrt{37}}\]
Unit vector that is normal to graph is \[\overrightarrow{\left( \dfrac{6}{\sqrt{37}},\dfrac{1}{\sqrt{37}} \right)}\] at (3,9).

Note: We should have a proper knowledge in slope and tangents for solving this type of question. This question will be easier for us to solve if we have a better knowledge of vectors. Also, remember that the value of multiplication of slope of two perpendicular lines is always -1.