What is the tangent of f(x)=6x+10-19e^x,(0,-9) ?

The tangent of f(x)=6x+10-19e^x,(0,-9) is y=-13x-9

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tangent of f(x)=6x+10-19e^x,(0,-9)

tangent of

f (x) = 6 x + 10 - 19 e^{x}, (0, - 9)

y = - 13 x - 9

Solution steps

Find the slope of

f (x) = 6 x + 10 - 19 e^{x} : \frac{df ( x )}{d x} = - 19 e^{x} + 6

Compute the slope of

f (x) = 6 x + 10 - 19 e^{x}

(0, - 9) : m = - 13

Find the line with slope m=

- 13

and passing through

(0, - 9) : y = - 13 x - 9

y = - 13 x - 9

\int \frac{9 x e ^{2 x}}{( 1 + 2 x ) ^{2}} d x

\int_{0}^{1} \frac{2 x}{( x ^{2} + 4 ) ^{2}} d x

x, \frac{1}{4} x

y^{^{'''}} - y^{^{''}} = 0

\frac{\partial}{\partial x} (2 x^{2} y + x e^{y^{2}})

What is the tangent of f(x)=6x+10-19e^x,(0,-9) ?
The tangent of f(x)=6x+10-19e^x,(0,-9) is y=-13x-9