What is the tangent of f(x)=2x^2-9x+10,\at x=2 ?

The tangent of f(x)=2x^2-9x+10,\at x=2 is y=-x+2

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tangent of f(x)=2x^2-9x+10,\at x=2

tangent of

f (x) = 2 x^{2} - 9 x + 10, at x = 2

y = - x + 2

Solution steps

Find the tangent point:

(2, 0)

Find the slope of

f (x) = 2 x^{2} - 9 x + 10 : \frac{df ( x )}{d x} = 4 x - 9

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 = - 1

Find the line with slope m=

- 1

and passing through

(2, 0) : y = - x + 2

y = - x + 2

\frac{d}{d x} (x arcsin (x))

tan (1 + e^{2 t})

\frac{\partial}{\partial x} (ln (5 x^{2} + 4 y^{2} + 9))

\int \frac{12 x ^{3}}{2 - x ^{4}} d x

f (x) = 5 x^{4} + \frac{1}{( 2 x ^{2} )}

What is the tangent of f(x)=2x^2-9x+10,\at x=2 ?
The tangent of f(x)=2x^2-9x+10,\at x=2 is y=-x+2