What is the tangent of f(x)=4x^2+5x-5 ?

The tangent of f(x)=4x^2+5x-5 is y=(8a_{0}+5)x-4a_{0}^2-5

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tangent of f(x)=4x^2+5x-5

tangent of

f (x) = 4 x^{2} + 5 x - 5

y = (8 a_{0} + 5) x - 4 a_{0}^{2} - 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{2} + 5 a_{0} - 5)

Find the slope of

f (x) = 4 x^{2} + 5 x - 5 : \frac{df ( x )}{d x} = 8 x + 5

Compute the slope of

f (x) = 4 x^{2} + 5 x - 5

x = a_{0} : m = 8 a_{0} + 5

Find the line with slope m=

8 a_{0} + 5

and passing through

(a_{0}, 4 a_{0}^{2} + 5 a_{0} - 5) : y = (8 a_{0} + 5) x - 4 a_{0}^{2} - 5

y = (8 a_{0} + 5) x - 4 a_{0}^{2} - 5

n = 1 \sum \infty \frac{( - 1 ) ^{n + 1}}{n ^{5}}

3 x - 2 x

n = 1 \sum \infty \frac{1}{cos ( n )}

\int - 6 sin (2 x) d x

\int_{0}^{3} x 11 - x^{2} d x

What is the tangent of f(x)=4x^2+5x-5 ?
The tangent of f(x)=4x^2+5x-5 is y=(8a_{0}+5)x-4a_{0}^2-5