What is the tangent of y=3x^2-4x+5 ?

The tangent of y=3x^2-4x+5 is y=(6a_{0}-4)x-3a_{0}^2+5

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tangent of y=3x^2-4x+5

tangent of

y = 3 x^{2} - 4 x + 5

y = (6 a_{0} - 4) x - 3 a_{0}^{2} + 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 a_{0}^{2} - 4 a_{0} + 5)

Find the slope of

y = 3 x^{2} - 4 x + 5 : \frac{d y}{d x} = 6 x - 4

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 = 6 a_{0} - 4

Find the line with slope m=

6 a_{0} - 4

and passing through

(a_{0}, 3 a_{0}^{2} - 4 a_{0} + 5) : y = (6 a_{0} - 4) x - 3 a_{0}^{2} + 5

y = (6 a_{0} - 4) x - 3 a_{0}^{2} + 5

What is the tangent of y=3x^2-4x+5 ?
The tangent of y=3x^2-4x+5 is y=(6a_{0}-4)x-3a_{0}^2+5