Find the slope of the line tangent to the curve f(x) = -10x^2 + 8x when x=9

The tangent line to f(x) at some point x has slope equal to f'(x), so first compute the derivative of f :

f(x) = -10x ² + 8x

f'(x) = -20x + 8

When x = 9, we have

f (9) = -10•9² + 8•9 = -738

f' (9) = -20•9 + 8 = -172

so the slope is -172.

If you also have to find the equation of the tangent line, use the point-slope formula:

y - (-738) = -172 (x - 9)

y = -172x + 810

region b will make the inequalities y < 2x + 1 and y > -x - 2 true ⇒ answer a

step-by-step explanation:

* lets study the graph and the inequality to solve the question

- the inequality y > -x - 2

∵ the form of the inequality is y > mx + c , m is the slope of the line

  and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = -2 that means the line intersect the y-axis at point (0 , -2)

∵ the sign of the inequality is >

∴ we shad the part over the line

  region a and b

- the inequality y < 2x + 1

∵ the form of the inequality is y > mx + c , m is the slope of the line

  and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = 1 that means the line intersect the y-axis at point (0 , 1)

∵ the sign of the inequality is <

∴ we shad the part under the line

  region b and c

∵ the common region between the two lines is region b

∴ region b will make the inequalities y < 2x + 1 and y > -x - 2 true