What is the tangent of y=5-7x^2,(3,-58) ?

The tangent of y=5-7x^2,(3,-58) is y=-42x+68

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tangent of y=5-7x^2,(3,-58)

tangent of

y = 5 - 7 x^{2}, (3, - 58)

y = - 42 x + 68

Solution steps

Find the slope of

y = 5 - 7 x^{2} : \frac{d y}{d x} = - 14 x

Compute the slope of

y = 5 - 7 x^{2}

(3, - 58) : m = - 42

Find the line with slope m=

- 42

and passing through

(3, - 58) : y = - 42 x + 68

y = - 42 x + 68

\int \frac{5 sin ( x )}{cos ^{3} ( x )} d x

\int \frac{x ^{2}}{2} d x

\int_{0}^{π} sin^{2} (2 x) d x

x \to 4 lim (3 - x)

y^{^{'}} = \frac{x ^{4}}{y}

What is the tangent of y=5-7x^2,(3,-58) ?
The tangent of y=5-7x^2,(3,-58) is y=-42x+68