# How do you find the equation of the tangent and normal line to the curve y=x^2+2x+3 at x=1?

Mar 8, 2017

a. Equation of tangent, $y = 4 x + 2$
b. Equation of normal, $4 y = - x + 25$

#### Explanation:

$y = {x}^{2} + 2 x + 3$

at $x = 1$, $y = {1}^{2} + 2 \left(1\right) + 3 = 6$

let say ${m}_{1}$ = gradient of tangent.
$\frac{d y}{d x} = 2 x + 2$

at $x = 1$, ${m}_{1} = \frac{d y}{d x} = 2 \left(1\right) + 2 = 4$

Therefore the equation of tangent at (1,6),
$\left(y - 6\right) = {m}_{1} \left(x - 1\right)$
$\left(y - 6\right) = 4 \left(x - 1\right)$
$y = 4 x - 4 + 6$
$y = 4 x + 2$

let say ${m}_{2}$ = gradient of normal.
${m}_{1} \cdot {m}_{2} = - 1$
${m}_{2} = - \frac{1}{4}$

Therefore the equation of normal at (1,6),
$\left(y - 6\right) = {m}_{2} \left(x - 1\right)$
$\left(y - 6\right) = - \frac{1}{4} \left(x - 1\right)$

$y = - \frac{1}{4} x + \frac{1}{4} + 6$

$y = - \frac{1}{4} x + \frac{25}{4}$

$4 y = - x + 25$