# Find an equation of the line that is tangent to the graph of f(x) = (x^2-4)^5 (2x-4)^3 for x...

## Question:

Find an equation of the line that is tangent to the graph of {eq}f(x) = (x^2-4)^5 (2x-4)^3 {/eq} for {eq}x = 1 {/eq}

## Equation of a Tangent Line:

The tangent line of a graph of a function at a point is simply the line that is parallel to the direction of the curve at the point, which also touches the curve at the point. To find the slope of the tangent line we find the derivative at that point.

We first find the derivative of the function by using the product rule and the chain rule:

{eq}\begin{align*} f'(x)& =( (x^2-4)^5)' (2x-4)^3 + (x^2-4)^5( (2x-4)^3)' \mbox{ (Product Rule)}\\ & = ( 5(x^2-4)^4) \cdot (x^2-4)' \cdot (2x-4)^3 + (x^2-4)^5( 3(2x-4)^2) \cdot (2x-4)' \mbox{ (Chain Rule)} \\ & = ( 5(x^2-4)^4) \cdot (2x)\cdot (2x-4)^3 + (x^2-4)^5( 3(2x-4)^2) \cdot 2 \end{align*} {/eq}

So at x=1, we have

{eq}\begin{align*} f'(1) &= ( 5(1^2-4)^4) \cdot (2\cdot 1)\cdot (2(1)-4)^3 + ((1)^2-4)^5( 3(2(1)-4)^2) \\& = ( 5(-3)^4) \cdot (2)\cdot (-2)^3 + (-3)^5( 3(-2)^2) \\ & = -9396 \end{align*} {/eq}

Note that the y value when x = 1 is:

{eq}f(1) = (1^2-4)^5 (2(1)-4)^3 = (-3)^5 (-2)^3 = 1944 {/eq}

So the equation of the tangent line at x=1 is:

{eq}y - 1944 = -9396(x-1) \\ y -1944 = -9396x + 9396 \\ y = -9396x +11340 {/eq} 