Slopes of Tangents (Leaving Cert Mathematics): Flashcards

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Slopes of Tangents
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Derivative for tangent slope

First derivative

What f′(x)f'(x) represents

Slope for any xx

Find slope at specific xx

Substitute xx into f′(x)f'(x)

Needs for tangent line equation

Slope and coordinate

Point-slope form

y−y1=m(x−x1)y-y_1=m(x-x_1)

Find yy for tangent coordinate

Substitute xx into original function

f′(x)f'(x) if f(x)=3x2−2x3f(x)=3x^2-2x^3

6x−6x26x-6x^2

f′(12)f'(\tfrac{1}{2}) if f′(x)=6x−6x2f'(x)=6x-6x^2

32\tfrac{3}{2}

yy at x=12x=\tfrac{1}{2} for f(x)=3x2−2x3f(x)=3x^2-2x^3

12\tfrac{1}{2}

Tangent eq. at (12,12)(\tfrac{1}{2},\tfrac{1}{2}) with m=32m=\tfrac{3}{2}

y=32x−14y=\tfrac{3}{2}x-\tfrac{1}{4}

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