Question

homework4: problem 12
(1 point)
let $f(x)=3x(5 - 4x)^2$.
find the equation of line tangent to the graph of $f$ at $x=-1$
tangent line: $y = $
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Explanation:

Step1: Expand the function

First, expand \(f(x)=3x(5 - 4x)^{2}=3x(25 - 40x+16x^{2}) = 75x-120x^{2}+48x^{3}\).

Step2: Find the derivative

Using the power - rule \((x^n)^\prime=nx^{n - 1}\), \(f^\prime(x)=75-240x + 144x^{2}\).

Step3: Find the slope of the tangent line

Evaluate \(f^\prime(x)\) at \(x=-1\). \(f^\prime(-1)=75+240 + 144=459\).

Step4: Find the y - coordinate of the point

Evaluate \(f(x)\) at \(x = - 1\). \(f(-1)=3\times(-1)\times(5 + 4)^{2}=-3\times81=-243\).

Step5: Use the point - slope form

The point - slope form of a line is \(y - y_1=m(x - x_1)\), where \((x_1,y_1)=(-1,-243)\) and \(m = 459\). So \(y+243=459(x + 1)\), which simplifies to \(y=459x+459 - 243=459x+216\).

Answer:

\(y = 459x+216\)