What is the tangent of f(x)=-3x^2-5x+1 ?

The tangent of f(x)=-3x^2-5x+1 is y=(-6a_{0}-5)x+3a_{0}^2+1

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tangent of f(x)=-3x^2-5x+1

tangent of

f (x) = - 3 x^{2} - 5 x + 1

y = (- 6 a_{0} - 5) x + 3 a_{0}^{2} + 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 3 a_{0}^{2} - 5 a_{0} + 1)

Find the slope of

f (x) = - 3 x^{2} - 5 x + 1 : \frac{df ( x )}{d x} = - 6 x - 5

Compute the slope of

f (x) = - 3 x^{2} - 5 x + 1

x = a_{0} : m = - 6 a_{0} - 5

Find the line with slope m=

- 6 a_{0} - 5

and passing through

(a_{0}, - 3 a_{0}^{2} - 5 a_{0} + 1) : y = (- 6 a_{0} - 5) x + 3 a_{0}^{2} + 1

y = (- 6 a_{0} - 5) x + 3 a_{0}^{2} + 1

\int cos^{4} (t) d t

y d y = 4 x (y^{2} + 1)^{\frac{1}{2}} d x

\int \frac{7 x ^{3} + 4 x}{x - 1} d x

y^{^{''}} + 25 y = - 40 sin (5 t)

x^{2} y^{^{'}} - 2 x y = y^{4}

What is the tangent of f(x)=-3x^2-5x+1 ?
The tangent of f(x)=-3x^2-5x+1 is y=(-6a_{0}-5)x+3a_{0}^2+1