What is the tangent of f(x)=26-x^2,(-5,1) ?

The tangent of f(x)=26-x^2,(-5,1) is y=10x+51

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tangent of f(x)=26-x^2,(-5,1)

tangent of

f (x) = 26 - x^{2}, (- 5, 1)

y = 10 x + 51

Solution steps

Find the slope of

f (x) = 26 - x^{2} : \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 = 10

Find the line with slope m=

10

and passing through

(- 5, 1) : y = 10 x + 51

y = 10 x + 51

f (x) = 2 x e^{- x}

\int \frac{1}{2 x + 1} d x

f (x) = \frac{1}{2 x - 5}

\frac{d}{d x} (2 x^{2})

3 x + 5

What is the tangent of f(x)=26-x^2,(-5,1) ?
The tangent of f(x)=26-x^2,(-5,1) is y=10x+51