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

The tangent of f(x)=4x^2-3x-7,\at x=2 is y=13x-23

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tangent of f(x)=4x^2-3x-7,\at x=2

tangent of

f (x) = 4 x^{2} - 3 x - 7, at x = 2

y = 13 x - 23

Solution steps

Find the tangent point:

(2, 3)

Find the slope of

f (x) = 4 x^{2} - 3 x - 7 : \frac{df ( x )}{d x} = 8 x - 3

Compute the slope of

f (x) = 4 x^{2} - 3 x - 7

x = 2 : m = 13

Find the line with slope m=

13

and passing through

(2, 3) : y = 13 x - 23

y = 13 x - 23

\frac{2 s}{( s ^{2} + 4 s + 13 )}

x \to 0 lim (sin (x) \frac{1}{x})

\int_{1}^{2} \frac{15}{x ^{3} + 2 x} d x

\frac{d}{d x} (2 x ln (x + 1))

\int \frac{( ln ( z ) ) ^{5}}{z} d z

What is the tangent of f(x)=4x^2-3x-7,\at x=2 ?
The tangent of f(x)=4x^2-3x-7,\at x=2 is y=13x-23