(a) find the slope m of the tangent to the curve y = 5 + 5x2 − 2x3 at the point where x =
a. m = (b) find equations of the tangent lines at the points (1, 8) and (2, 9). y(x) = (at the point (1, 8)) y(x) = (at the point (2, 9))

a) The slope at (1, 8) is 4.
The slope at (2, 9) is -4.
These values are computed numerically by the graphing calculator and shown in the column f'(x) in the graphic.

The derivative is y' = 10x -6x². At x=1, y' = 10-6 = 4; at x=2, y' = 20-24 = -4.

b) In point-slope form, the equations of the tangent lines are shown in the graphic.

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Cricetus Cricetus · Answer 2 · 2021-09-29T09:50:20+02:00

The slope of the tangent will be "[tex]10a-6a^2[/tex]". The equation at point (1, 8) is "[tex]y = 4x+4[/tex]" and at point (2,9) is "[tex]y = -4x+17[/tex]".

Given equation is:

[tex]y = 5+5x^2-2x^3[/tex]

(a)

The slope of the tangent to the curve point where [tex]x = a[/tex] is:

→ [tex]\frac{dy}{dx} = m = 10x-6x^2[/tex]

→ [tex]= 10a-6a^2[/tex]

(b)

The equation of tangent at point (1,8) will be:

→ Slope, [tex]m = 10-6[/tex]

[tex]= 4[/tex]

→ [tex]y-8=4(x-1)[/tex]

[tex]y = 4x-4+8[/tex]

[tex]y = 4x+4[/tex]

and,

The equation of tangent at point (2,9) will be:

→ Slope, [tex]m = 10\times 2-6\times 2^2[/tex]

[tex]= 20-24[/tex]

[tex]= -4[/tex]

→ [tex]y-9=-4(x-2)[/tex]

[tex]y-9=-4x+8[/tex]

[tex]y = -4x+17[/tex]

Thus the above answer is right.

Learn more about tangent here:

https://brainly.com/question/10462284

(a) find the slope m of the tangent to the curve y = 5 + 5x2 − 2x3 at the point where x = a. m = (b) find equations of the tangent lines at the points (1, 8) and (2, 9). y(x) = (at the point (1, 8)) y(x) = (at the point (2, 9))

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(a) find the slope m of the tangent to the curve y = 5 + 5x2 − 2x3 at the point where x =
a. m = (b) find equations of the tangent lines at the points (1, 8) and (2, 9). y(x) = (at the point (1, 8)) y(x) = (at the point (2, 9))