What is the tangent of y=38x-16x^2,\at x=1 ?

The tangent of y=38x-16x^2,\at x=1 is y=6x+16

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tangent of y=38x-16x^2,\at x=1

tangent of

y = 38 x - 16 x^{2}, at x = 1

y = 6 x + 16

Solution steps

Find the tangent point:

(1, 22)

Find the slope of

y = 38 x - 16 x^{2} : \frac{d y}{d x} = 38 - 32 x

Compute the slope of

y = 38 x - 16 x^{2}

x = 1 : m = 6

Find the line with slope m=

6

and passing through

(1, 22) : y = 6 x + 16

y = 6 x + 16

\int \frac{16}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

- 4 x + \frac{1}{x} - 3

\frac{\partial}{\partial x} (x^{2} + 2 y + 2 z - 12)

\frac{10}{( s + 1 ) ^{2} ( s + 3 )}

\int \frac{1}{169 - x ^{2}} d x

What is the tangent of y=38x-16x^2,\at x=1 ?
The tangent of y=38x-16x^2,\at x=1 is y=6x+16