What is the tangent of f(x)=4x^2+3x,(-3,27) ?

The tangent of f(x)=4x^2+3x,(-3,27) is y=-21x-36

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tangent of f(x)=4x^2+3x,(-3,27)

tangent of

f (x) = 4 x^{2} + 3 x, (- 3, 27)

y = - 21 x - 36

Solution steps

Find the slope of

f (x) = 4 x^{2} + 3 x : \frac{df ( x )}{d x} = 8 x + 3

Compute the slope of

f (x) = 4 x^{2} + 3 x

(- 3, 27) : m = - 21

Find the line with slope m=

- 21

and passing through

(- 3, 27) : y = - 21 x - 36

y = - 21 x - 36

x \to \infty lim (2 x - x^{2})

\frac{\partial}{\partial x} (e^{2 x yz - 3})

y^{^{'}} = 9 y^{2}

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

\frac{d y}{d x} = (y (x y^{3} - 1))

What is the tangent of f(x)=4x^2+3x,(-3,27) ?
The tangent of f(x)=4x^2+3x,(-3,27) is y=-21x-36