What is the tangent of f(x)=0.14x^2+0.68x+3.1 ?

The tangent of f(x)=0.14x^2+0.68x+3.1 is y=(0.28a_{0}+0.68)x-0.14a_{0}^2+3.1

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tangent of f(x)=0.14x^2+0.68x+3.1

tangent of

f (x) = 0.14 x^{2} + 0.68 x + 3.1

y = (0.28 a_{0} + 0.68) x - 0.14 a_{0}^{2} + 3.1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 0.14 a_{0}^{2} + 0.68 a_{0} + 3.1)

Find the slope of

f (x) = 0.14 x^{2} + 0.68 x + 3.1 : \frac{df ( x )}{d x} = 0.28 x + 0.68

Compute the slope of

f (x) = 0.14 x^{2} + 0.68 x + 3.1

x = a_{0} : m = 0.28 a_{0} + 0.68

Find the line with slope m=

0.28 a_{0} + 0.68

and passing through

(a_{0}, 0.14 a_{0}^{2} + 0.68 a_{0} + 3.1) : y = (0.28 a_{0} + 0.68) x - 0.14 a_{0}^{2} + 3.1

y = (0.28 a_{0} + 0.68) x - 0.14 a_{0}^{2} + 3.1

y = sin (4 x), (0, \frac{π}{4})

y^{^{'}} = 2 x + y, y (0) = 5

\frac{d}{d x} (- e^{x})

\int cot^{4} (x) csc^{4} (x) d x

\int_{0}^{\frac{π}{2}} 5 cos^{2} (x) d x

What is the tangent of f(x)=0.14x^2+0.68x+3.1 ?
The tangent of f(x)=0.14x^2+0.68x+3.1 is y=(0.28a_{0}+0.68)x-0.14a_{0}^2+3.1