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

The tangent of f(x)=2x^2-3,(2,5) is y=8x-11

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

tangent of

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

y = 8 x - 11

Solution steps

Find the slope of

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

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

Find the line with slope m=

8

and passing through

(2, 5) : y = 8 x - 11

y = 8 x - 11

What is the tangent of f(x)=2x^2-3,(2,5) ?
The tangent of f(x)=2x^2-3,(2,5) is y=8x-11