How to find the $y$-intercept ($b$) from a tangent line

Question

The original function is given, $$y = 2x^2+2x+7$$ at the point $(4,47)$.

I found the tangent line from the original function which is: $$f'(x)=4x+2.$$ The question states that the tangent line can be written as $y=mx+b$ and asks me to find the slope ($m$) as well as the $y$-intercept ($b$).

I have found the slope which is $18$, by plugging in the $x$-value from the point $(4,47)$ into the tangent line: so I have $f(x) = 4x+2$ which equals $4(4)+2$ which is $18$.

The way I'm trying to find the $y$-intercept is by doing: $y-47=18(x-3)$ which is $-7$, but it's wrong the answer is $-25$.

What mistake am I making that's leading me to get $-7$?

Answer 1

Instead of $y-47=18(x-3)$, you should say $y-\color{blue}{47}=18(x-\color{red}4)$,

because the point $(\color{red}4,\color{blue}{47})$ is on the tangent line,

and then you'll get the correct answer.

Answer 2

You're making a mistake at this part : $$x-\color{cyan}3$$ it should be $$x-\color{cyan}4$$You can also use $y = mx+c$ instead of the complicated $y-y1 = m(x-x1)$. Because you already know $m$ and a point $(4,47)$ then you just need to find $c$, that way is gonna be easier and will make you avoid calculation mistake