What is the tangent of f(x)=3x^2-5x+8,\at x=2 ?

The tangent of f(x)=3x^2-5x+8,\at x=2 is y=7x-4

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tangent of f(x)=3x^2-5x+8,\at x=2

tangent of

f (x) = 3 x^{2} - 5 x + 8, at x = 2

y = 7 x - 4

Solution steps

Find the tangent point:

(2, 10)

Find the slope of

f (x) = 3 x^{2} - 5 x + 8 : \frac{df ( x )}{d x} = 6 x - 5

Compute the slope of

f (x) = 3 x^{2} - 5 x + 8

x = 2 : m = 7

Find the line with slope m=

7

and passing through

(2, 10) : y = 7 x - 4

y = 7 x - 4

\frac{2}{( s ^{2} - 1 ) ( s - 1 )}

\frac{\partial}{\partial y} (x^{2} + 5 y^{2})

x \to 3 - lim (- \frac{3}{7} x + \frac{4}{11})

\frac{\partial}{\partial z} (\frac{x}{y + z})

x \to 0 - lim (\frac{( x + 4 )}{( 3 )})

What is the tangent of f(x)=3x^2-5x+8,\at x=2 ?
The tangent of f(x)=3x^2-5x+8,\at x=2 is y=7x-4