Question

find an equation of the tangent line to the curve at the given point.
$y = 7x^2 - x^3$, $(1, 6)$
$y = \square$

Explanation:

Step1: Compute derivative of y

$y' = 14x - 3x^2$

Step2: Find slope at x=1

Substitute $x=1$: $y'(1)=14(1)-3(1)^2=11$

Step3: Use point-slope form

Formula: $y - y_1 = m(x - x_1)$
Substitute $(1,6)$ and $m=11$: $y - 6 = 11(x - 1)$
Simplify: $y = 11x - 5$

Answer:

$11x - 5$