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

The tangent of f(x)=x^2-7x+5,\at x=4 is y=x-11

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tangent of f(x)=x^2-7x+5,\at x=4

tangent of

f (x) = x^{2} - 7 x + 5, at x = 4

y = x - 11

Solution steps

Find the tangent point:

(4, - 7)

Find the slope of

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

Compute the slope of

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

x = 4 : m = 1

Find the line with slope m=

1

and passing through

(4, - 7) : y = x - 11

y = x - 11

\int \frac{2 x ^{2} + 6 x - 5 x}{3 x ^{2}} d x

y = - csc (x) - cos (x)

2 y^{^{''}} + 8 y = 0

\int_{- \infty}^{- 1} e^{2 x} d x

\int 6 x^{2} - 3 x^{3} d x

What is the tangent of f(x)=x^2-7x+5,\at x=4 ?
The tangent of f(x)=x^2-7x+5,\at x=4 is y=x-11