What is the tangent of 6-5x^2 ?

The tangent of 6-5x^2 is y=-10a_{0}x+6+5a_{0}^2

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tangent of 6-5x^2

tangent of

6 - 5 x^{2}

y = - 10 a_{0} x + 6 + 5 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 6 - 5 a_{0}^{2})

Find the slope of

y = 6 - 5 x^{2} : \frac{d y}{d x} = - 10 x

Compute the slope of

y = 6 - 5 x^{2}

x = a_{0} : m = - 10 a_{0}

Find the line with slope m=

- 10 a_{0}

and passing through

(a_{0}, 6 - 5 a_{0}^{2}) : y = - 10 a_{0} x + 6 + 5 a_{0}^{2}

y = - 10 a_{0} x + 6 + 5 a_{0}^{2}

n = 1 \sum \infty \frac{5}{n ^{2} + 9}

\int_{\frac{1}{5}}^{3} 6 x ln (5 x) d x

\frac{2 x}{x - 1}

y^{^{'}} = \frac{1}{( t - 1 ) ( y + 1 )}, y (0) = 1

\int \frac{1}{s ^{2} - 2 s + 5} d s