What is the tangent of f(x)=2x^2-7x+5,\at x=3 ?

The tangent of f(x)=2x^2-7x+5,\at x=3 is y=5x-13

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tangent of f(x)=2x^2-7x+5,\at x=3

tangent of

f (x) = 2 x^{2} - 7 x + 5, at x = 3

y = 5 x - 13

Solution steps

Find the tangent point:

(3, 2)

Find the slope of

f (x) = 2 x^{2} - 7 x + 5 : \frac{df ( x )}{d x} = 4 x - 7

Compute the slope of

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

x = 3 : m = 5

Find the line with slope m=

5

and passing through

(3, 2) : y = 5 x - 13

y = 5 x - 13

\int_{0}^{t} cos (x^{2}) d x

\frac{\partial}{\partial x} (cos (4 θ))

\int \frac{sin ^{2} ( x )}{1 + cos ( x )} d x

y = x + cos (x)

\frac{d}{d x} (\frac{x}{1})

What is the tangent of f(x)=2x^2-7x+5,\at x=3 ?
The tangent of f(x)=2x^2-7x+5,\at x=3 is y=5x-13