The graph of the function is $g(x)=x^3-2x^2+x+1$ and the tangent to the curve is at $x=2$?

Question

a) Find the equation of the tangent to the curve at $x=2$

HELP

and then

b) Determine the angle that this tangent makes with the positive direction of the $x$-axis

Please help I really need to know how to do this Please include working

Thanks.

For part a) I found the gradient by doing $g'(x)$ and subbing in $2$ for $x$ and I got $5$ so, so far I have $y=5x+c$, dunno how to find $c$ though

Answer 1

$$ f(2)=2^3-3\cdot2^2 + 2 + 1 = -1. $$ So the point $(2,-1)$ is on the graph. Therefore you need the line that passes through that point and whose slope is $5$.

Answer 2

For the first $x_0=2\to y_0=g(2)=\cdots, \ m=g'(2)=\cdots\implies y-y_0=m(x-x_0)$. Or complete your solution using the point $(x_0,y_0)$
The angle $\theta$ satisfies $m=\tan\theta$

Answer 3

When $y=x^3-2x^2+x+1$ then $$m_{\text{tanget}}=y'(x)|_{x=2}=3(2^2)-4(2)+1=5$$ so we get this line $$y-y(2)=m(x-2)\to y=1+5(x-2)$$

Answer 4

(a) Ok so we have the equation $$f(x) = x^3 - 2x^2 + x + 1$$ Taking the derivative we get $$f'(x) = 3x^2 - 4x + 1$$ At $x = 2$, $f'(2) = 3*4 - 4*2 + 1 = 5$, meaning that our equation of the tangent line is $$y = 5x + c$$ $f(2) = 8 - 2*4 + 2 + 1 = 3$, so the graph passes through the point $(2, 3)$. Our tangent line equation is now $$y = 5x -7$$ (b) so since the graph passes through the point (0, -7) and (7/5, 0), we have a triangle with a height of 7 and a base of 7/5. You can find theta with that information.