Find the slope of the line tangent tangent to the graph a the given point. Y = x

Question

Find the slope of the line tangent tangent to the graph a the given point. Y = x^2 + 6x, x = -8 m = -16 m = 16 m = -2 m = -10 Find an equation for the tangent to the curve at the given point. Y =x^2 - 1, (-3, 8) y = -3x - 10 y = -6x - 10 y = -6x - 19 y = -6x - 20 Calculate the derivative of the function. The n find the value of the derivative as specified. F(x) = 5x + 9; f'(2) f'(x) = 5x; f'(2) = 10 f'(x) = 0; f'(2) = 0 f'(x) = 5; f'(2) = 5 f'(x) = 9; f'(2) = 9 find the derivative. Y = x^14 14x^14 14x^13 13x^14 13x^13 y = 12 11 12 1 0 y = - 2x^7 - 2x^6 - 14x^8 - 14x^6 - 14x^7 y = 1/3 x^6 18x^5 2x^5 1/18 x^6 2x^7

Explanation / Answer

1). The function is y=x2+6x so the slope y'=2x and at x=-8, m=2*-8=-16. So the option is A) m=-16

2) Equation of tangent is y-y1= m(x-x1). The slope y'=2x and at x=-3 m=2*-3=-6. So the equation of tangent is

y-8=-6(x+3) which on simplification gives y= -6x-10

3) Given that f(x)=5x+9. f'(x)=5(1)+0; f'(x)=5. As there is no expression of x the value of f'(2) also is five so the correct option is C) f'(x)=5 and f'(2)=5

4) The derivative for y=x14 y'=14x14-1= 14x13. So the correct option is B)

5) y=12 The derivative for constant is always zero So y'=0

6) y= -2x7. The derivative y'= -2*7x7-1= -14x6. The correct option is C)

7) y= 1/3 x6. y'=1/3 * 6 x6-1= 2x5. The correct option is B)

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