What is the tangent of y=49-x^2,(-7,0) ?

The tangent of y=49-x^2,(-7,0) is y=14x+98

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tangent of y=49-x^2,(-7,0)

tangent of

y = 49 - x^{2}, (- 7, 0)

y = 14 x + 98

Solution steps

Find the slope of

y = 49 - x^{2} : \frac{d y}{d x} = - 2 x

Compute the slope of

y = 49 - x^{2}

(- 7, 0) : m = 14

Find the line with slope m=

14

and passing through

(- 7, 0) : y = 14 x + 98

y = 14 x + 98

\frac{\partial}{\partial x} (y^{2} - x \cdot y)

\int ln (2 - x) d x

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

\frac{d ^{2}}{d x ^{2}} (x^{\frac{5}{2}} e^{x})

What is the tangent of y=49-x^2,(-7,0) ?
The tangent of y=49-x^2,(-7,0) is y=14x+98