What is the tangent of f(x)=2x^2+5x-3 ?

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

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tangent of f(x)=2x^2+5x-3

tangent of

f (x) = 2 x^{2} + 5 x - 3

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

f (x) = 2 x^{2} + 5 x - 3 : \frac{df ( x )}{d x} = 4 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 = 4 a_{0} + 5

Find the line with slope m=

4 a_{0} + 5

and passing through

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

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

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