What is the tangent of y=8x-x^2,(1,7) ?

The tangent of y=8x-x^2,(1,7) is y=6x+1

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tangent of y=8x-x^2,(1,7)

tangent of

y = 8 x - x^{2}, (1, 7)

y = 6 x + 1

Solution steps

Find the slope of

y = 8 x - x^{2} : \frac{d y}{d x} = 8 - 2 x

Compute the slope of

y = 8 x - x^{2}

(1, 7) : m = 6

Find the line with slope m=

6

and passing through

(1, 7) : y = 6 x + 1

y = 6 x + 1

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