tangent of (x-2)(x^2+4) — step-by-step solution | Explain MAths

tangent of (x-2)(x^2+4)

No x-coordinate is specified: find the tangent at the general point

x = a

, for real

a

⇥Let's differentiate:

\frac{d}{d x} (x^{3} - 2 x^{2} + 4 x - 8)

⇥

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

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

+ \frac{d}{d x} (4 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} (4 x)

=

4

⇥

3 x^{2} - 2 \cdot 2 x + 4

⇥

3 x^{2} - 4 x + 4

The point of tangency is (

a

,

a^{3}

- 2 a^{2}

+ 4 a

- 8

)

The slope is f'(

a

) =

3 a^{2} - 4 a + 4

The tangent passes through this point with this slope:

y = (3 a^{2} - 4 a + 4) \cdot (x - a) + (a^{3}

- 2 a^{2}

+ 4 a

- 8)

⇥

(a^{3}

- 2 a^{2}

+ 4 a

- 8) - (3 a^{2} - 4 a + 4) \cdot a

⇥

a^{3}

- 2 a^{2}

+ 4 a

- 8

- (3 a^{2} - 4 a + 4) \cdot a

⇥

a^{3}

- 2 a^{2}

+ 4 a

- 8

- (3 a^{3} - 4 a^{2} + 4 a)

⇥

a^{3}

- 2 a^{2}

+ 4 a

- 8

- 3 a^{3}

+ 4 a^{2}

- 4 a

⇥ℹ

a^{3} - 3 a^{3}

=

-2 a^{3}

‖

- 2 a^{2} + 4 a^{2}

=

2 a^{2}

‖

4 a - 4 a

=

0

⇥

- 2 a^{3}

+ 2 a^{2}

+ 0

- 8

⇥

- 2 a^{3} + 2 a^{2} - 8

⇥

y = (3 a^{2} - 4 a + 4) \cdot x + (- 2 a^{3} + 2 a^{2} - 8)

▶Tangent line:

y = (3 a^{2} - 4 a + 4) \cdot x

- 2 a^{3}

+ 2 a^{2}

- 8