1. ## Simple tangent/normal question

The curve C has equation y = x² - 4x + 7. The point A has coordinates (1,4).

a] Find the equation of the tangent to C at A.

b] Find the equation of the normal to C at the point A.

Now I'm pretty sure I differentiate the equation first to get 2x-4, but then I have no idea at all what to do after that. How is this done?

2. Originally Posted by db5vry
The curve C has equation y = x² - 4x + 7. The point A has coordinates (1,4).

a] Find the equation of the tangent to C at A.

b] Find the equation of the normal to C at the point A.

Now I'm pretty sure I differentiate the equation first to get 2x-4, but then I have no idea at all what to do after that. How is this done?

$\displaystyle f(x)=x^2-4x+7$ and $\displaystyle f'(x)=2x-4$

Then the slope of the tangent in A is

$\displaystyle m_t=f'(4)~\implies~m_t = 4$

Then the perpendicular direction (the direction of the normal) is:

$\displaystyle m_{\perp}=-\dfrac1{m_t} ~\implies~m_{\perp} = -\dfrac14$

Use the point-slope-formula to determine the equation of the tangent:

$\displaystyle y-4=4(x-1)~\implies~y=4x$

Use the point-slope-formula to determine the equation of the normal:

$\displaystyle y-4=-\dfrac14(x-1)~\implies~y=-\dfrac14 x + \dfrac{17}4$