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Find the equation of the tangent line to $k(x) = x^2$ at $x = 6$ . $\newline$ Write your answer in point-slope form using integers and fractions. Simplify any fractions. $\newline$ $y - \underline{\quad} = \underline{\quad}(x - \underline{\quad})$

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Find equations of tangent lines using limits

Full solution

Q. Find the equation of the tangent line to

k(x) = x^2

at

x = 6

.

\newline

Write your answer in point-slope form using integers and fractions. Simplify any fractions.

\newline

y - \underline{\quad} = \underline{\quad}(x - \underline{\quad})

Find Derivative of $k(x)$ : Find the derivative of $k(x)$ to determine the slope of the tangent line at $x = 6$ .
$k(x) = x^2$
$k'(x) = 2x$
$k'(6) = 2(6) = 12$
Calculate Slope and Point: Use the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$ , where $m$ is the slope and $(x_1, y_1)$ is a point on the line. $\newline$ We already know the slope $m = 12$ from Step $1$ . $\newline$ To find $y_1$ , substitute $x = 6$ into $k(x)$ : $\newline$ $k(6) = 6^2 = 36$ $\newline$ So, the point $(6, 36)$ lies on the tangent line.

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