tangent of x^3-2x^2-5x+6,\at x=2 — step-by-step solution

⚠Ⓘ 2x is read as the product 2*x: for a power, write 2^x‖5x is read as the product 5*x: for a power, write 5^x

1) Find the tangent at

x = 2

⇥1.1) Let's differentiate:

\frac{d}{d x} (x^{3} - 2 x^{2} - 5 x + 6)

⇥

\frac{d}{d x} (x^{3})

+ \frac{d}{d x} (- 2 x^{2})

+ \frac{d}{d x} (- 5 x)

+ 0

⇥ℹ

\frac{d}{d x} (x^{3})

3 x^{2}

‖

\frac{d}{d x} (- 2 x^{2})

-2 \cdot \frac{d}{d x} (x^{2})

‖

\frac{d}{d x} (x^{2})

2 x

‖

\frac{d}{d x} (- 5 x)

-5

⇥

3 x^{2} - 2 \cdot 2 x - 5

⇥

3 x^{2} - 4 x - 5

Compute f(

2

)

⇥

2^{3}

- 2 \cdot 2^{2}

- 5 \cdot 2

+ 6

⇥ℹ

2^{3}

8

‖

2^{2}

4

‖

- 5 \cdot 2

-10

⇥

8

- 2 \cdot 4

- 10

+ 6

⇥ℹ

- 2 \cdot 4

-8

‖

8 - 10

-2

⇥

- 2 - 8 + 6

⇥

- 10 + 6

⇥

-4

The point of tangency is (

2

-4

)

Compute f'(

2

)

⇥

3 \cdot 2^{2} - 4 \cdot 2 - 5

⇥ℹ

2^{2}

4

‖

- 4 \cdot 2

-8

⇥

3 \cdot 4 - 8 - 5

⇥ℹ

3 \cdot 4

12

‖

- 8 - 5

-13

⇥

12 - 13

⇥

-1

The slope is f'(

2

) =

-1

The tangent passes through this point with this slope:

y = - 1 \cdot (x - 2) - 4

Compute the y-intercept:

b = - 4 - (-1 \cdot 2)

⇥

- 4 - (-1 \cdot 2)

⇥

- 4 + 2

⇥

-2

⇥

y = - 1 \cdot x - 2

▶Tangent line:

y = - x - 2