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

Step 1: Determine the general equation for the slope of the tangent
The slope of a line is given by the derivative of the function.
Given `f(x)=x^2+5`
the slope is `(df(x))/(dx)=2x` (using the exponent rule for exponents)

Step 2: Determine the specific slope of the tangent at the given point
At `(3,9)`, `x=3`
So the slope is `(df(3))/(dx)=2 * 3=6`

Step 3: Determine the unit vector with slope of the tangent
Consider a unit vector with slope of `6` and a base at the origin:
`color(white)("XXX")y/x=6`

`color(white)("XXX")rarr y = 6x`

and since it is a unit vector:
`color(white)("XXX")sqrt(x^2+y^2)=1`

`color(white)("XXX")rarrsqrt(x^2+(6x)^2)=1`

`color(white)("XXX")rarr sqrt(37)x=1`

`color(white)("XXX")rarr x=1/sqrt(37)`

`color(white)("XXX")rarr y=6/sqrt(37)`

Unit vector parallel to the tangent: `(vec(x,y)) = (vec(1/37,6/37))`

Step 4: Determine the unit vector perpendicular to the tangent
Remember that if a line has a slope of `m` then all lines perpendicular to it will have a slope of `-1/m`.
So we are looking for a unit vector with slope `(-1/6)`
Using the same logic as in Step 3, we have
`color(white)("XXX")(vec(x,y))=(vec(1/sqrt(37),-6/sqrt(37)))`