What is the tangent of f(x)=x^2+3x+5,(1,9) ?

The tangent of f(x)=x^2+3x+5,(1,9) is y=5x+4

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tangent of f(x)=x^2+3x+5,(1,9)

tangent of

f (x) = x^{2} + 3 x + 5, (1, 9)

y = 5 x + 4

Solution steps

Find the slope of

f (x) = x^{2} + 3 x + 5 : \frac{df ( x )}{d x} = 2 x + 3

Compute the slope of

f (x) = x^{2} + 3 x + 5

(1, 9) : m = 5

Find the line with slope m=

5

and passing through

(1, 9) : y = 5 x + 4

y = 5 x + 4

\int arccot (10 x) d x

\frac{\partial}{\partial y} (e^{(x + y)})

\int \frac{x}{2 x + 7} d x

\frac{\partial}{\partial x} (5 cos (7 x - 3 y))

x \to 2 + lim (\frac{1}{∣ 2 - x ∣})

What is the tangent of f(x)=x^2+3x+5,(1,9) ?
The tangent of f(x)=x^2+3x+5,(1,9) is y=5x+4