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

The tangent of f(x)=x^2+x-3,\at x=4 is y=9x-19

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

tangent of

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

y = 9 x - 19

Solution steps

Find the tangent point:

(4, 17)

Find the slope of

f (x) = x^{2} + x - 3 : \frac{df ( x )}{d x} = 2 x + 1

Compute the slope of

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

x = 4 : m = 9

Find the line with slope m=

9

and passing through

(4, 17) : y = 9 x - 19

y = 9 x - 19

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What is the tangent of f(x)=x^2+x-3,\at x=4 ?
The tangent of f(x)=x^2+x-3,\at x=4 is y=9x-19