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Ind the slope m of the tangent to the curve y = 9 + 4x2 − 2x3 at the point where x =
a. m = 8a−6a2 (b) find equations of the tangent lines at the points (1, 11) and (2, 9). y(x) = 2x+11 (at the point (1, 11)) y(x) = (at the point (2, 9))

Respuesta :

Find the derivative:-
y' = 8x - 6x^2
This is the slope in terms of x.
When x = a , the slope is
8a - 6a^2

(b)  equation of tangent line at x1,y1 
is  y- y1  =  (8x - 6x^2)( x - x1) , so at (1,11) it is
y - 11 = (8(1) - 6((1)^2) ) (x - 1)
y =  2(x - 1) + 11
y = 2x + 9 (answer)

At  (2,9) 
y - 9 = (16-24)(x - 2)
y - 9 = -8x + 16
y = -8x + 25  answer

This is about slope of tangents via derivatives.

A) m = 8a- 6a²

B) At (1, 11), equation of tangent is y = 2x + 9

At (2, 9), equation of tangent is y = 2x + 4

  • The slope of a tangent to a curve is simply gotten by finding the first derivative of the function representing that curve.

  • We are given the function; y = 9 + 4x² - 2x³

The slope will be;

m = y' = 8x - 6x²

A) At x = a; To find the slope we will just put a for x in the slope function;

Slope; m = 8a- 6a²

B) Slope is 8x - 6x²

At the point (1, 11);

Slope = 8(1) - 6(1²) = 2

Thus, equation of tangent line is; y - 11 = 2(x - 1)

⇒ y - 11 = 2x - 2

⇒ y = 2x - 2 + 11

⇒ y = 2x + 9

  • At the point (2, 9);

Slope = 8(2) - 6(2²) = -8

Thus, equation of tangent line is; y - 8 = 2(x - 2)

⇒ y - 8 = 2x - 4

⇒ y = 2x - 4 + 8

⇒ y = 2x + 4

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