Tangent to $y=(1+2x)^2$ at $(4,81)$

Question

Find the equation of the tangent line to the curve $y=(1+2x)^2$ at the point $(4,81)$.

Answer 1

The derivative of the curve is $y'=8x+4$, which at point $(4, 81)$ is equal to $8(4)+4 = 36$, which tells us the slope of the tangent line.

Therefore $y = 36x + b$ is the equation of the tangent line in general. At point $(4, 81)$, we have $81 = 36(4) + b$ or $b=-63$.

So the equation of the tangent line is $y = 36x - 63$.

Answer 2

Well, the derivative is $$y'=2(1+2x)2=4(1+2x)$$ giving the slope $m=4(1+8)=36$ and hence the required line is $$y-81=36(x-4)$$

Answer 3

we have the line in the form $$y=mx+n$$ where $m$ in the slope in the given point and in $n$ intersects the line the $y$-axes. $$y'(x)=4(1+2x)$$ thus $$y'(4)=4(1+8)=36$$ and we have $$y=36x+n$$ inserting $x=4,y=81$ and we get $n$ $$81=36\cdot 4+n$$

Answer 4

We have, $y={(1+2x)}^2$

Thus, $\frac{dy}{dx} = 2.2(1+2x)$

At $(4,81)$, $\frac{dy}{dx} = 36$

Now, you have a point and the slope, finding the equation of a line is easy.

Answer 5

If you are in a calculus class, you should read your text or notes with an eye towards understanding what they are saying. They must be telling you that the equation of the line tangent to the graph of $y=f(x)$ at $(x_0,f(x_0))$ is $$y=y_0+(x-x_0)f'(x_0)$$

This follows directly from the facts that (1) if $(x,y)$ is another point on the tangent line, then the slope of the tangent is $$\frac{y-y_0}{x-x_0}$$ and (2) the slope of the tangent line is $$f'(x_0)$$ Equating these two gives you an equation which you can easily rewrite in the form I stated above.