What is the tangent of f(x)=2x^2-7x+5,\at x=1 ?

The tangent of f(x)=2x^2-7x+5,\at x=1 is y=-3x+3

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tangent of f(x)=2x^2-7x+5,\at x=1

tangent of

f (x) = 2 x^{2} - 7 x + 5, at x = 1

y = - 3 x + 3

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

f (x) = 2 x^{2} - 7 x + 5 : \frac{df ( x )}{d x} = 4 x - 7

Compute the slope of

f (x) = 2 x^{2} - 7 x + 5

x = 1 : m = - 3

Find the line with slope m=

- 3

and passing through

(1, 0) : y = - 3 x + 3

y = - 3 x + 3

y = 5 x^{3} - 2 x^{4}, x = 0, x = 2

\frac{d}{d x} (6 x^{2} \cdot cos (2 x))

\frac{d}{d t} (e^{t s i n (2 t)})

\frac{d}{d x} (9 e^{- x})

\frac{d}{d θ} (5 + 5 sin (θ))

What is the tangent of f(x)=2x^2-7x+5,\at x=1 ?
The tangent of f(x)=2x^2-7x+5,\at x=1 is y=-3x+3