What is the tangent of y=10x-x^2,(5,26) ?

The tangent of y=10x-x^2,(5,26) is y=2x+16 \lor y=-2x+36

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tangent of y=10x-x^2,(5,26)

tangent of

y = 10 x - x^{2}, (5, 26)

y = 2 x + 16 or y = - 2 x + 36

Solution steps

Slope of the tangent line at

(5, 26) : m = \frac{10 x - x ^{2} - 26}{x - 5}

Find the slope of

y = 10 x - x^{2} : m = 10 - 2 x

Solve

10 - 2 x = \frac{10 x - x ^{2} - 26}{x - 5} : x = 4, x = 6

Find the tangent line at

x = 4 : y = 2 x + 16

Find the tangent line at

x = 6 : y = - 2 x + 36

y = 2 x + 16 or y = - 2 x + 36

\int \frac{1}{x - ln ^{2} ( x ) + ln ( x ) + 3} d x

\frac{d}{d t} (2 e^{t} + 2 e^{2 t})

\int \frac{2 x + 5}{9 - x ^{2}} d x

y = sin (e^{2 x})

\frac{d}{d x} (\frac{x + 3}{8 x ^{2} e ^{x}})

What is the tangent of y=10x-x^2,(5,26) ?
The tangent of y=10x-x^2,(5,26) is y=2x+16 \lor y=-2x+36