What is the tangent of y= 8/(1-t),(-1,4) ?

The tangent of y= 8/(1-t),(-1,4) is y=2t+6

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tangent of y= 8/(1-t),(-1,4)

tangent of

y = \frac{8}{1 - t}, (- 1, 4)

y = 2 t + 6

Solution steps

Find the slope of

y = \frac{8}{1 - t} : \frac{d y}{d t} = \frac{8}{( 1 - t ) ^{2}}

Compute the slope of

y = \frac{8}{1 - t}

(- 1, 4) : m = 2

Find the line with slope m=

2

and passing through

(- 1, 4) : y = 2 t + 6

y = 2 t + 6

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

y = 9 x - 2 x^{2}, (- 2, - 26)

\frac{d}{d x} (\frac{6 x}{6 + x})

\frac{\partial}{\partial x} (\frac{( x + y )}{y})

y = - 0.0959 x^{2} + 106.83 x + 71149

What is the tangent of y= 8/(1-t),(-1,4) ?
The tangent of y= 8/(1-t),(-1,4) is y=2t+6