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

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

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

tangent of

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

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

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} + 2)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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} + 2) : y = (- 8 a_{0} - 5) x + 4 a_{0}^{2} + 2

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

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