What is the tangent of f(x)=1-7x^2,(-2,-27) ?

The tangent of f(x)=1-7x^2,(-2,-27) is y=28x+29

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tangent of f(x)=1-7x^2,(-2,-27)

tangent of

f (x) = 1 - 7 x^{2}, (- 2, - 27)

y = 28 x + 29

Solution steps

Find the slope of

f (x) = 1 - 7 x^{2} : \frac{df ( x )}{d x} = - 14 x

Compute the slope of

f (x) = 1 - 7 x^{2}

(- 2, - 27) : m = 28

Find the line with slope m=

28

and passing through

(- 2, - 27) : y = 28 x + 29

y = 28 x + 29

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

y^{^{''}} + 13 y^{^{'}} + 42 y = 0

y^{^{''}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = 0

\int \frac{x ^{6}}{6} d x

\int_{2}^{8} x d x

What is the tangent of f(x)=1-7x^2,(-2,-27) ?
The tangent of f(x)=1-7x^2,(-2,-27) is y=28x+29