What is the tangent of y=x^2+5x+6,\at x=7 ?

The tangent of y=x^2+5x+6,\at x=7 is y=19x-43

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tangent of y=x^2+5x+6,\at x=7

tangent of

y = x^{2} + 5 x + 6, at x = 7

y = 19 x - 43

Solution steps

Find the tangent point:

(7, 90)

Find the slope of

y = x^{2} + 5 x + 6 : \frac{d y}{d x} = 2 x + 5

Compute the slope of

y = x^{2} + 5 x + 6

x = 7 : m = 19

Find the line with slope m=

19

and passing through

(7, 90) : y = 19 x - 43

y = 19 x - 43

\int 144 - 9 x^{2} d x

x \to - 5 + lim (\frac{3}{x ^{2} - 25})

\frac{d}{d x} (x^{s i n (3 x)})

\frac{\partial}{\partial u} (cos (u + v))

x \to - 4 + lim (\frac{1}{( x + 4 ) ^{2}})

What is the tangent of y=x^2+5x+6,\at x=7 ?
The tangent of y=x^2+5x+6,\at x=7 is y=19x-43