What is the tangent of y=x^2+5x+3,\at x=1 ?

The tangent of y=x^2+5x+3,\at x=1 is y=7x+2

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tangent of y=x^2+5x+3,\at x=1

tangent of

y = x^{2} + 5 x + 3, at x = 1

y = 7 x + 2

Solution steps

Find the tangent point:

(1, 9)

Find the slope of

y = x^{2} + 5 x + 3 : \frac{d y}{d x} = 2 x + 5

Compute the slope of

y = x^{2} + 5 x + 3

x = 1 : m = 7

Find the line with slope m=

7

and passing through

(1, 9) : y = 7 x + 2

y = 7 x + 2

\frac{d}{d x} (3 (5 - 7 x)^{5})

\frac{\partial}{\partial x} (x^{\frac{3}{4}} y^{- 4})

\int x sin^{2} (x) d x

\int (x^{2} - 1) e^{2 x} d x

y^{^{''}} - 6 y^{^{'}} + 4 y = 0

What is the tangent of y=x^2+5x+3,\at x=1 ?
The tangent of y=x^2+5x+3,\at x=1 is y=7x+2