What is the tangent of f(x)=9x^2+7x-6,\at x=-7/18 ?

The tangent of f(x)=9x^2+7x-6,\at x=-7/18 is y=-265/36

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tangent of f(x)=9x^2+7x-6,\at x=-7/18

tangent of

f (x) = 9 x^{2} + 7 x - 6, at x = - \frac{7}{18}

y = - \frac{265}{36}

Solution steps

Find the tangent point:

(- \frac{7}{18}, - \frac{265}{36})

Find the slope of

f (x) = 9 x^{2} + 7 x - 6 : \frac{df ( x )}{d x} = 18 x + 7

Compute the slope of

f (x) = 9 x^{2} + 7 x - 6

x = - \frac{7}{18} : m = 0

Find the line with slope m=

0

and passing through

(- \frac{7}{18}, - \frac{265}{36}) : y = - \frac{265}{36}

y = - \frac{265}{36}

\frac{d}{d x} (1 - ∣ x ∣)

\int e^{x} + cos (x) d x

\int \frac{5 x + 3}{7 x - 1} d x

\frac{y}{5}

\frac{d ^{2}}{d x ^{2}} (12 x^{3} - 32 x^{2})

What is the tangent of f(x)=9x^2+7x-6,\at x=-7/18 ?
The tangent of f(x)=9x^2+7x-6,\at x=-7/18 is y=-265/36