What is the tangent of f(x)=4x^2-5x(-1.9) ?

The tangent of f(x)=4x^2-5x(-1.9) is y=(8a_{0}+9.5)x-4a_{0}^2

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tangent of f(x)=4x^2-5x(-1.9)

tangent of

f (x) = 4 x^{2} - 5 x (- 1.9)

y = (8 a_{0} + 9.5) x - 4 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{2} + 9.5 a_{0})

Find the slope of

f (x) = 4 x^{2} - 5 x (- 1.9) : \frac{df ( x )}{d x} = 8 x + 9.5

Compute the slope of

f (x) = 4 x^{2} - 5 x (- 1.9)

x = a_{0} : m = 8 a_{0} + 9.5

Find the line with slope m=

8 a_{0} + 9.5

and passing through

(a_{0}, 4 a_{0}^{2} + 9.5 a_{0}) : y = (8 a_{0} + 9.5) x - 4 a_{0}^{2}

y = (8 a_{0} + 9.5) x - 4 a_{0}^{2}

\int_{3}^{4} \frac{1}{( x - 3 ) ^{2}} d x

(\frac{3}{5}, \frac{2}{3}), (- \frac{3}{5}, \frac{3}{4})

\int (e^{t})^{2} d t

\int x^{3} sec^{2} (y) d y

y^{^{''}} + \frac{y ^{^{'}}}{x} = 2 x + 1

What is the tangent of f(x)=4x^2-5x(-1.9) ?
The tangent of f(x)=4x^2-5x(-1.9) is y=(8a_{0}+9.5)x-4a_{0}^2