Question

Derivative/graphing tangent lines help?

Answer 1

5) The function f(x) = - 2x²  + 2x + 3

y = - 2x²  + 2x + 3

For x = 1, y = - 2(1)²  + 2(1) + 3

y = - 2 + 2 + 3

y = 3

Tangent point (1, 3)

y = - 2x²  + 2x + 3

Differentiating  on each side with respect to x .

y' = - 4x + 2

Substitute x = 1

y' = - 4(1) + 2

y' = - 4 + 2

y' = - 2

This is the slope of tangent line to the curve at (1, 3).

To find the tangent line equation, substitute the values of m = - 2 and (x, y ) = (1, 3)  in the slope intercept form of an equation y = mx + b.

3 = - 2(1) + b

3 + 2 = b

b = 5

Substitute m = -2 and b = 5 in y = mx + b.

y = - 2x + 5

The above equation in standard form ax + by = c is 2x + y = 5

Tangent line equation is 2x + y = 5.

Option d is correct.

Answer 2

6) The function f(x) = x²  - 2

y = x²  - 2

Differentiating  on each side with respect to x .

y' = 2x

Here y' is slope of the tangent line.

2x  = 3

x = 3/2

Using the x - value find the corresponding y - value with the curve.

y = (3/2)²  - 2

y = (9/4) - 2

y = (9 - 8)/4

y = 1/4

Therefore, the point is (3/2, 1/4).

Option e is correct.

Answer 3

7) The function f(x) = - x³ + 3x²  - 2

y = - x³ + 3x²  - 2

Differentiating on each side with respect to x .

y' = - 3x² + 6x

Here y' is slope of the tangent line.

Horizontal tangent line slope is zero.

Equate y' = 0

- 3x² + 6x = 0

Divide each side by 3.

- x² + 2x = 0

Take out common factor.

- x( x - 2) = 0

Apply zero product property.

- x = 0 and x - 2 = 0

x = 0 and x = 2

Using the x - values find the corresponding y - value with the curve.

For x = 0, y = - (0)³ + 3(0)²  - 2

y = - 2

For x = 2, y = - (2)³ + 3(2)²  - 2

y = - 8 + 12 - 2

y = 2

The points are (0, -2) and (2, 2).

Option e is correct.