What is the tangent of f(x)=x^2-5,(3,4) ?

The tangent of f(x)=x^2-5,(3,4) is y=6x-14

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tangent of f(x)=x^2-5,(3,4)

tangent of

f (x) = x^{2} - 5, (3, 4)

y = 6 x - 14

Solution steps

Find the slope of

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

Compute the slope of

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

(3, 4) : m = 6

Find the line with slope m=

6

and passing through

(3, 4) : y = 6 x - 14

y = 6 x - 14

y = \frac{1}{x ^{7}}

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

h \to 0 lim (\frac{( h + 2 ) ^{2} - 4}{h})

y = \frac{2 x}{x + 5}, (5, 1)

\frac{d ^{2}}{d x ^{2}} (3 x^{2} - 5 y^{2})

What is the tangent of f(x)=x^2-5,(3,4) ?
The tangent of f(x)=x^2-5,(3,4) is y=6x-14