What is the tangent of 2x^2-9x+10,\at x=1 ?

The tangent of 2x^2-9x+10,\at x=1 is y=-5x+8

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tangent of 2x^2-9x+10,\at x=1

tangent of

2 x^{2} - 9 x + 10, at x = 1

y = - 5 x + 8

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

y = 2 x^{2} - 9 x + 10 : \frac{d y}{d x} = 4 x - 9

Compute the slope of

y = 2 x^{2} - 9 x + 10

x = 1 : m = - 5

Find the line with slope m=

- 5

and passing through

(1, 3) : y = - 5 x + 8

y = - 5 x + 8

\int \frac{( 1 - 2 y )}{y} d y

\int e^{2 x} \cdot sin (2 x) d x

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

(\frac{1}{s}) (\frac{10}{s + 5})

\frac{3 x + 4}{2 x - 1}

What is the tangent of 2x^2-9x+10,\at x=1 ?
The tangent of 2x^2-9x+10,\at x=1 is y=-5x+8