What is the tangent of f(x)=4x^2-4x+1,(-1,8) ?

The tangent of f(x)=4x^2-4x+1,(-1,8) is y=-8x \lor y=-16x-8

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tangent of f(x)=4x^2-4x+1,(-1,8)

tangent of

f (x) = 4 x^{2} - 4 x + 1, (- 1, 8)

y = - 8 x or y = - 16 x - 8

Solution steps

Slope of the tangent line at

(- 1, 8) : m = \frac{4 x ^{2} - 4 x - 7}{x + 1}

Find the slope of

f (x) = 4 x^{2} - 4 x + 1 : m = 8 x - 4

Solve

8 x - 4 = \frac{4 x ^{2} - 4 x - 7}{x + 1} : x = - \frac{1}{2}, x = - \frac{3}{2}

Find the tangent line at

x = - \frac{1}{2} : y = - 8 x

Find the tangent line at

x = - \frac{3}{2} : y = - 16 x - 8

y = - 8 x or y = - 16 x - 8

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What is the tangent of f(x)=4x^2-4x+1,(-1,8) ?
The tangent of f(x)=4x^2-4x+1,(-1,8) is y=-8x \lor y=-16x-8