What is the tangent of f(x)= 7/2 x^2-5x+1 ?

The tangent of f(x)= 7/2 x^2-5x+1 is y=(7a_{0}-5)x-7/2 a_{0}^2+1

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tangent of f(x)= 7/2 x^2-5x+1

tangent of

f (x) = \frac{7}{2} x^{2} - 5 x + 1

y = (7 a_{0} - 5) x - \frac{7}{2} a_{0}^{2} + 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{7}{2} a_{0}^{2} - 5 a_{0} + 1)

Find the slope of

f (x) = \frac{7}{2} x^{2} - 5 x + 1 : \frac{df ( x )}{d x} = 7 x - 5

Compute the slope of

f (x) = \frac{7}{2} x^{2} - 5 x + 1

x = a_{0} : m = 7 a_{0} - 5

Find the line with slope m=

7 a_{0} - 5

and passing through

(a_{0}, \frac{7}{2} a_{0}^{2} - 5 a_{0} + 1) : y = (7 a_{0} - 5) x - \frac{7}{2} a_{0}^{2} + 1

y = (7 a_{0} - 5) x - \frac{7}{2} a_{0}^{2} + 1

θ \to 0 lim (θ cot (5 θ))

\frac{d}{d x} (\frac{x ^{2} - 9}{x ^{2} + 9})

x^{^{''}} + x^{^{'}} + 2 x = 0

\frac{d y}{d x} = 2 y (3 - y)

y = 5 arctan (x - 1 + x^{2})

What is the tangent of f(x)= 7/2 x^2-5x+1 ?
The tangent of f(x)= 7/2 x^2-5x+1 is y=(7a_{0}-5)x-7/2 a_{0}^2+1