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

The slope of the tangent line to the graph of $f(x) = x^2 + 3$ at any point $x$ is given by the derivative of $f(x)$, which is $f'(x) = 2x$. At $x = 3$, the slope is $f'(3) = 2(3) = 6$. The equation of the tangent line can be found using the point-slope form: $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is the point of tangency. Substituting $m = 6$ and $(x_1, y_1) = (3, 12)$, we get $y - 12 = 6(x - 3)$, which simplifies to $y = 6x - 6$.

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 $f ( x ) = x ^ { 2 } + 3$