What is the tangent of y=-3x^2+10x+2,\at x=1 ?

The tangent of y=-3x^2+10x+2,\at x=1 is y=4x+5

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tangent of y=-3x^2+10x+2,\at x=1

tangent of

y = - 3 x^{2} + 10 x + 2, at x = 1

y = 4 x + 5

Solution steps

Find the tangent point:

(1, 9)

Find the slope of

y = - 3 x^{2} + 10 x + 2 : \frac{d y}{d x} = - 6 x + 10

Compute the slope of

y = - 3 x^{2} + 10 x + 2

x = 1 : m = 4

Find the line with slope m=

4

and passing through

(1, 9) : y = 4 x + 5

y = 4 x + 5

\frac{t d u}{d t} = t^{2} + 3 u, t > 0, H (6) = 180

5 - 2 x^{2}

\frac{d y}{d x} = 0.05 x - 450

\frac{0.6}{s + 10}

\int \frac{e ^{x} - 3}{x} d x

What is the tangent of y=-3x^2+10x+2,\at x=1 ?
The tangent of y=-3x^2+10x+2,\at x=1 is y=4x+5