tangent of 5x^2+8x-8 — step-by-step solution

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

x = a

, for real

a

Let's differentiate:

\frac{d}{d x} (5 x^{2} + 8 x - 8)

ⓘ The derivative of a sum is the sum of the derivatives:

\frac{d}{d x} (5 x^{2} + 8 x - 8)

\frac{d}{d x} (5 x^{2}) + \frac{d}{d x} (8 x) + 0

\frac{d}{d x} (5 x^{2}) + \frac{d}{d x} (8 x) + 0

ⓘ A constant factor stays in front of the derivative, (k*u)' = k*u':

\frac{d}{d x} (5 x^{2})

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

\frac{d}{d x} (8 x)

8

■The derivative of x^n is n*x^(n-1):

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

2 x

■Adding 0 changes nothing:

\frac{d}{d x} (8 x) + 0

\frac{d}{d x} (8 x)

5 \cdot 2 x + 8

ⓘ Multiplications are done before additions and subtractions:

5 \cdot 2

10

10 x + 8

Compute f(

a

)

The point of tangency is (

a

5 a^{2} + 8 a - 8

)

Compute f'(

a

)

The slope is f'(

a

) =

10 a + 8

The tangent passes through this point with this slope:

y = (10 a + 8) \cdot (x - a) + (5 a^{2} + 8 a - 8)

Compute the y-intercept:

b = (5 a^{2} + 8 a - 8) - (10 a + 8) \cdot a

(5 a^{2} + 8 a - 8) - (10 a + 8) \cdot a

5 a^{2}

+ 8 a

- 8

- (10 a + 8) \cdot a

ⓘ Expand by multiplying each term inside the parentheses:

- (10 a + 8) \cdot a

- (10 a^{2} + 8 a)

5 a^{2}

+ 8 a

- 8

- (10 a^{2} + 8 a)

ⓘ A minus sign before a parenthesis changes the sign of each of its terms:

- (10 a^{2} + 8 a)

- 10 a^{2} - 8 a

5 a^{2}

+ 8 a

- 8

- 10 a^{2}

- 8 a

ⓘ Like terms are grouped by adding their coefficients:

5 a^{2} - 10 a^{2}

-5 a^{2}

8 a - 8 a

0

- 5 a^{2} + 0 - 8

ⓘ Adding 0 changes nothing:

0 - 8

-8

- 5 a^{2} - 8

y = (10 a + 8) \cdot x + (- 5 a^{2} - 8)

▶Tangent line:

y = (10 a + 8) \cdot x - 5 a^{2} - 8