Find the equation of the tangent lone to the curve y=x^3+4x^2+4 at (1,9)

Question

accessdenied · Answer

1) Find the derivative of y with respect to x
2) Plug in the x-value of point into derivative equation for slope of line at x
3) Use point-slope form to find line equation
  y - y1 = m(x - x1)

rogue · Answer

You can use the point slope form, or you can use the linearization form, which I find more efficient. The tangent line to the curve f(x) at (Xo, Yo is$$Y_{tangent} = f'(x_0)(x - x_0 )+y_0$$

anonymous · Answer

would the answer be y=11x-2

rogue · Answer

Yup, good job! =)

anonymous · Answer

would you be able to show me how to do it step by step?

rogue · Answer

Yeah, sure, hold on.

rogue · Answer

Tangent line to the curve y=x^3+4x^2+4 at (1,9).

Step 1: Find y'(1)$$ y=x^3+4x^2+4$$$$y' = \frac {d}{dx} \left[ x^3+4x^2+4 \right]$$$$y' = 3x^2 + 8x$$$$y' (1) = 3 * 1^2 + 8 *1 = 11$$
Step 2: Plug in your info. into tangent line equation. You have f'(1) = 11, the x and y coordinates the line passes to be x = 1, and y = 9.$$Y_{tangent} = f'(x_0)(x - x_0 )+y_0$$$$Y_{tangent} = 11*(x - 1 )+9 = 11x - 11 + 9 = 11x - 2$$
Done! Does that help?

anonymous · Answer

thank you i did it correct really appreciate it (:

rogue · Answer

Alrighty, awesome :)