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

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

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

tangent of

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

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

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

x = a_{0} : m = 4 a_{0} - 3

Find the line with slope m=

4 a_{0} - 3

and passing through

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

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

\int \frac{x ^{2} + 2 x + 3}{x ^{3} + 3 x ^{2} + 9 x} d x

y = x^{2} - 2 x + 3, [- 2, 2]

\int_{0}^{1} \frac{7}{1 + t ^{2}} d t

\frac{d}{d x} (6 + 2 ln (x))

\frac{d}{d x} (x - 5 x^{2})

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