What is the tangent of-x^2+9x+3,\at x=5 ?

The tangent of-x^2+9x+3,\at x=5 is y=-x+28

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tangent of-x^2+9x+3,\at x=5

tangent of

- x^{2} + 9 x + 3, at x = 5

y = - x + 28

Solution steps

Find the tangent point:

(5, 23)

Find the slope of

y = - x^{2} + 9 x + 3 : \frac{d y}{d x} = - 2 x + 9

Compute the slope of

y = - x^{2} + 9 x + 3

x = 5 : m = - 1

Find the line with slope m=

- 1

and passing through

(5, 23) : y = - x + 28

y = - x + 28

y = x^{2}, y = x, y = 0

n = 2 \sum \infty e^{3 - 4 n}

f (x) = (2 x - 8) 4 (x^{2} + x + 1) 5

\frac{s + 2}{s ^{2} - s - 12}

x \to 0 lim (3 x^{7 x})

What is the tangent of-x^2+9x+3,\at x=5 ?
The tangent of-x^2+9x+3,\at x=5 is y=-x+28