How do I find the tangent of an equation, when the $x$-coordinate is $1$? Find the equation of the tangent to the graph $y = x^2 + 4x + 3$, whose $x$ coordinate is $1$.
Write your answer in form $y = mx + c$
I got the equation $y = 2x^2 + 2x + 2$.
Is this correct?

$$y=x^2+4x+3$$
$$\frac{dy}{dx}=2x+4$$
At $x=1$, y is 8 and the derivative is 6, so write
$y=mx+c\rightarrow8=6(1)+c$; we get $c=2$. Hence the tangent at $x=1$ is $y=6x+2$.
