I'm trying to determine the tangents to the curve $\displaystyle f(x)=2x^2+3$ that pass through the point $\displaystyle (2,3)$

I got the first equation right, but the second one is apparently wrong (according to the book).

What I did was set the derivative equal to the slope of two points. Points I used were $\displaystyle (x,f(x))$ and $\displaystyle (2,3)$ which which would leave the slope to be $\displaystyle (2x^2)/(2-3)$

$\displaystyle 4x=(2x^2)/(2-3)$

$\displaystyle 0=2x(3x-4)$

$\displaystyle x=0, 4/3$

First equation is the tangent that passes through $\displaystyle f(0)=3$ so $\displaystyle (0,3)$. Finding the equation by setting the derivative at $\displaystyle f(0)$ equal to the slope like this:

$\displaystyle 0=(y-3)/(x-0)$

$\displaystyle y=3$

But I'm getting a different answer for the second equation:/

The book has the second equation as: $\displaystyle 16x-y-29=o$