Find the equation of the tangent line to the graph of $f ( x ) = x ^ { 2 } + 3$ at the point (3,12).
A)y = 12
B)y = 6x - 6
C)x = 3
D)Not defined

Question

Quizplus · Accepted Answer

To find the equation of the tangent line to the graph of $f(x) = x^2 + 3$ at the point (3,12), we first need to find the derivative of $f(x)$, which gives us the slope of the tangent line at any point $x$. The derivative $f'(x) = 2x$. At $x = 3$, the slope is $f'(3) = 2*3 = 6$. Using the point-slope form 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, we get $y - 12 = 6(x - 3)$. Simplifying, we find the equation of the tangent line to be $y = 6x - 6$.

Find the Equation of the Tangent Line to the Graph f(x)=x2+3f ( x ) = x ^ { 2 } + 3f(x)=x2+3

Find the Equation of the Tangent Line to the Graph $f ( x ) = x ^ { 2 } + 3$