What is the tangent of f(x)=5x^2-4x+3 ?

The tangent of f(x)=5x^2-4x+3 is y=(10a_{0}-4)x-5a_{0}^2+3

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tangent of f(x)=5x^2-4x+3

tangent of

f (x) = 5 x^{2} - 4 x + 3

y = (10 a_{0} - 4) x - 5 a_{0}^{2} + 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 5 a_{0}^{2} - 4 a_{0} + 3)

Find the slope of

f (x) = 5 x^{2} - 4 x + 3 : \frac{df ( x )}{d x} = 10 x - 4

Compute the slope of

f (x) = 5 x^{2} - 4 x + 3

x = a_{0} : m = 10 a_{0} - 4

Find the line with slope m=

10 a_{0} - 4

and passing through

(a_{0}, 5 a_{0}^{2} - 4 a_{0} + 3) : y = (10 a_{0} - 4) x - 5 a_{0}^{2} + 3

y = (10 a_{0} - 4) x - 5 a_{0}^{2} + 3

y^{^{''}} = k^{2} \cdot y

\int 5 e^{- x} d x

y^{^{'}} - 4 x y = 0

y = 8 cos (x)

cos^{2} (x) sin (x) y^{^{'}} + (cos^{3} (x)) y = 1

What is the tangent of f(x)=5x^2-4x+3 ?
The tangent of f(x)=5x^2-4x+3 is y=(10a_{0}-4)x-5a_{0}^2+3