What is the tangent of 7x^2+4x-5 ?

The tangent of 7x^2+4x-5 is y=(14a_{0}+4)x-7a_{0}^2-5

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tangent of 7x^2+4x-5

tangent of

7 x^{2} + 4 x - 5

y = (14 a_{0} + 4) x - 7 a_{0}^{2} - 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 7 a_{0}^{2} + 4 a_{0} - 5)

Find the slope of

y = 7 x^{2} + 4 x - 5 : \frac{d y}{d x} = 14 x + 4

Compute the slope of

y = 7 x^{2} + 4 x - 5

x = a_{0} : m = 14 a_{0} + 4

Find the line with slope m=

14 a_{0} + 4

and passing through

(a_{0}, 7 a_{0}^{2} + 4 a_{0} - 5) : y = (14 a_{0} + 4) x - 7 a_{0}^{2} - 5

y = (14 a_{0} + 4) x - 7 a_{0}^{2} - 5

\int - 5 x cos (x) d x

\frac{d}{d x} ((y - x)^{2})

x \to 2 lim (∣ x - 3 ∣)

\int cos (t) - sin (t) d t

What is the tangent of 7x^2+4x-5 ?
The tangent of 7x^2+4x-5 is y=(14a_{0}+4)x-7a_{0}^2-5