What is the tangent of f(x)=-8x^2+10,(0.5,8) ?

The tangent of f(x)=-8x^2+10,(0.5,8) is y=-8x+12

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tangent of f(x)=-8x^2+10,(0.5,8)

tangent of

f (x) = - 8 x^{2} + 10, (0.5, 8)

y = - 8 x + 12

Solution steps

Find the slope of

f (x) = - 8 x^{2} + 10 : \frac{df ( x )}{d x} = - 16 x

Compute the slope of

f (x) = - 8 x^{2} + 10

(0.5, 8) : m = - 8

Find the line with slope m=

- 8

and passing through

(0.5, 8) : y = - 8 x + 12

y = - 8 x + 12

\frac{d}{d x} ((x^{2} - 9)^{\frac{1}{2}})

n = 1 \sum \infty (- \frac{2}{3})^{n}

\frac{d}{d x} (ln (sec (2 x)))

f (x) = \frac{4 π ^{2}}{6 - x}

\frac{d}{d x} ((x + x^{- 1})^{3})

What is the tangent of f(x)=-8x^2+10,(0.5,8) ?
The tangent of f(x)=-8x^2+10,(0.5,8) is y=-8x+12