equation of the normal?
y =(4x-1)(x+2)/2x

find the equation of the normal at the point when x = -2
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answers:
Steve4Physics say: y = (4x-1)(x+2)/(2x)
. .= (4x² + 7x – 2)/(2x)
. .= 2x + 7/2 – 1/x

dy/dx = 2 + 1/x²

When x=2, dy/dx = 2 + 1/2² = 2 + ¼ = 9/4

At x=2 the tangent has gradient m = 9/4, so the normal has gradient = -1/m = -4/9

When x = -2, y = (4x-1)(x+2)/(2x) = (4*(-2) - 1)(-2+2)/(2*-2) = 0

So the normal passes through (-2, 0) with gradient -4/9 so its equation is:
y – 0 = (-4/9)(x – (-2))
. . . . .= (-4/9)(x + 2)
. . . . .= (-4/9)x - 8/9
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alex say: y = 2x - 1/x + 7/2
y'=2 + 1/x^2
y'(-2) = 9/4
equation of the normal
y=-(4/9)(x+2)
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ffjfb say: lymdpppb
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hjffh say: qfsepbcc
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