What is the tangent of f(x)=x^2-5,\at x=1 ?

The tangent of f(x)=x^2-5,\at x=1 is y=2x-6

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tangent of f(x)=x^2-5,\at x=1

tangent of

f (x) = x^{2} - 5, at x = 1

y = 2 x - 6

Solution steps

Find the tangent point:

(1, - 4)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 2

Find the line with slope m=

2

and passing through

(1, - 4) : y = 2 x - 6

y = 2 x - 6

x^{3}, at \frac{16}{3}

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (1) = 0, y^{^{'}} (1) = 8

- cot (x)

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

\int \frac{x + 2}{x ^{2} + 4 x} d x

What is the tangent of f(x)=x^2-5,\at x=1 ?
The tangent of f(x)=x^2-5,\at x=1 is y=2x-6