Find the equation of the tangent and the normal to the following curves at the indicated points:
y = x2 + 4x + 1 at x = 3
finding slope of the tangent by differentiating the curve
m(tangent) at (3,0) = 10
normal is perpendicular to tangent so, m1m2 = – 1
equation of tangent is given by y – y1 = m(tangent)(x – x1)
y at x = 3
y = 32 + 4×3 + 1
y = 22
y – 22 = 10(x – 3)
y = 10x – 8
equation of normal is given by y – y1 = m(normal)(x – x1)
x + 10y = 223