What is the tangent of y=x^2,\at x=-3/4 ?

The tangent of y=x^2,\at x=-3/4 is y=-3/2 x-9/16

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tangent of y=x^2,\at x=-3/4

tangent of

y = x^{2}, at x = - \frac{3}{4}

y = - \frac{3}{2} x - \frac{9}{16}

Solution steps

Find the tangent point:

(- \frac{3}{4}, \frac{9}{16})

Find the slope of

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

Compute the slope of

y = x^{2}

x = - \frac{3}{4} : m = - \frac{3}{2}

Find the line with slope m=

- \frac{3}{2}

and passing through

(- \frac{3}{4}, \frac{9}{16}) : y = - \frac{3}{2} x - \frac{9}{16}

y = - \frac{3}{2} x - \frac{9}{16}

\frac{d}{d x} (ln ∣ csc (x) - cot (x) ∣)

y = x^{2} - 2 x, y = 2 - x

\int 3^{4 x + 6} d x

\frac{\partial}{\partial x} (2 y + x^{2} + 1)

2 cos^{2} (t)

What is the tangent of y=x^2,\at x=-3/4 ?
The tangent of y=x^2,\at x=-3/4 is y=-3/2 x-9/16