# `f(x) = (x - 2)(x^2 + 4), (1,-5)` (a) Find an equation of the tangent line to the graph of f at the given point, (b) use a graphing utility to graph the function and its tangent line at the...

`f(x) = (x - 2)(x^2 + 4), (1,-5)` (a) Find an equation of the tangent line to the graph of f at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results.

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The function `f(x) = (x - 2)/(x^2+4)`

The slope of the tangent at point x = a is f'(a).

`f'(x) = ((x - 2)'*(x^2+4) - (x - 2)(x^2+4)')/(x^2+4)^2`

= `((x^2+4) - (x - 2)(2x))/(x^2+4)^2`

= `(x^2+4 - 2x^2 +4x)/(x^2+4)^2`

= `(-x^2+4 +4x)/(x^2+4)^2`

At the point where x = 1, y = -1/5 not -5.

f'(1) = 7/25 = 0.28

The equation of the tangent at (1, -1/5) is `(y + 1/5)/(x - 1) = 7/25`

`(5y + 1)/(x - 1) = 7/5`

`25y + 5 = 7x - 7`

`y = (7x - 12)/25`

The graph of the function and the tangent at (1, -1/5) is:

Start by taking the derivative. This will require product rule and chain rule.

The product rule is: AB'+BA'

`f'(x) = (x-2)(2x) + (x^2+4)(1)`

`f'(x)= (x-2)(2x) + x^2+4`

`f'(x)= 2x^2 -4x+x^2+4`

The derivative is:

`f'(x) = 3x^2-4x+4`

Substitute the x-value of the given point.

`f'(1) = 3(1)^2-4(1)+4`

`f'(1)=3`

The slope at the point (1,-5) is 3.

The equation of the tangent line is in the form of:

`y=mx+b`

Substitute the slope, 3, and the given point, (1,-5), into the equation and find the y-intercept, b.

`-5=(3)(1)+b`

`-5=3+b`

`b=-8`

Substitute the slope and y-intercept back into the equation of the tangent line.

**a)**

`y=3x-8`

Graph the original function with the equation of the tangent line. See image attached.

**b) See the image below: ↓**