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

The tangent of f(x)=x^2+4x-5,\at x=2 is y=8x-9

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tangent of f(x)=x^2+4x-5,\at x=2

tangent of

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

y = 8 x - 9

Solution steps

Find the tangent point:

(2, 7)

Find the slope of

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

Compute the slope of

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

x = 2 : m = 8

Find the line with slope m=

8

and passing through

(2, 7) : y = 8 x - 9

y = 8 x - 9

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What is the tangent of f(x)=x^2+4x-5,\at x=2 ?
The tangent of f(x)=x^2+4x-5,\at x=2 is y=8x-9