What is the tangent of f(x)=x^2+1,(4,17) ?

The tangent of f(x)=x^2+1,(4,17) is y=8x-15

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tangent of f(x)=x^2+1,(4,17)

tangent of

f (x) = x^{2} + 1, (4, 17)

y = 8 x - 15

Solution steps

Find the slope of

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

Compute the slope of

f (x) = x^{2} + 1

(4, 17) : m = 8

Find the line with slope m=

8

and passing through

(4, 17) : y = 8 x - 15

y = 8 x - 15

h (x) = (x^{2} - 3 x + 2) (2 x^{3} + 1)

x^{3}, at x = - 1

\frac{\partial}{\partial x} (6 y x^{5} - 20 y^{2} x^{4})

y = x, y = 2 x

\frac{\partial}{\partial x} (cos (x) sin (y))

What is the tangent of f(x)=x^2+1,(4,17) ?
The tangent of f(x)=x^2+1,(4,17) is y=8x-15