Question

y = 3x^2 + 2x - 7 find slope of tangent and normal at point (3,2)

Answer 1

dy/dx=6x+2=m
m at (3,2)= 6 (3)+2=18+2=20
slope of tangent = 20
slope of normal =-1/20

Answer 2

$y = 3x^{2} + 2x - 7 \\ (3) \\ y = 3 ({3})^{2} + 2(3) - 7 \\ = 3(9) + 6 - 7 \\ = 27 - 1 \\ = 26 \\ \\ y = {3x}^{2} + 2x - 7 \\ = (2) \\ y = 3( 2)^{2} + 2 ({2)}^{2} - 7 \\ = 3(4) + 4 - 7 \\ = 12 - 3 \\ = 9 \\ \\ y = 26 - 9 \\ y = 17$