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

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Answer 1 · 2017-03-08T04:55:29

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

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

at $x =1$ , $y = 1^2+ 2(1) + 3 =6$

let say $m_1$ = gradient of tangent.
$(d y)/(d x) = 2 x + 2$

at $x =1$ , $m_1 = (d y)/(d x) = 2 (1) + 2 =4$

Therefore the equation of tangent at (1,6),
$(y-6) = m_1(x-1)$
$(y-6) = 4(x-1)$
$y =4 x -4 +6$
$y =4 x + 2$

let say $m_2$ = gradient of normal.
$m_1 * m_2 = -1$
$m_2 = -1/4$

Therefore the equation of normal at (1,6),
$(y-6) = m_2(x-1)$
$(y-6) = -1/4(x-1)$

$y = -1/4 x + 1/4 +6$

$y = -1/4 x + 25/4$

$4 y = -x +25$

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