The equation of the tangent to the parabola y | vedclass

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(D) Given the parabola equation $y = x^2 - x$.At $x = 1$,the $y$-coordinate is $y = (1)^2 - 1 = 0$. So,the point of tangency is $(1, 0)$.To find the s

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$y = x - 1$

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(D) Given the parabola equation $y = x^2 - x$.At $x = 1$,the $y$-coordinate is $y = (1)^2 - 1 = 0$. So,the point of tangency is $(1, 0)$.To find the slope of the tangent,we differentiate $y$ with respect to $x$: $\frac{dy}{dx} = 2x - 1$.Evaluating the derivative at $x = 1$: $\left( \frac{dy}{dx} \right)_{x=1} = 2(1) - 1 = 1$.Using the point-slope form of the line equation $y - y_1 = m(x - x_1)$,where $m = 1$ and $(x_1, y_1) = (1, 0)$:$y - 0 = 1(x - 1)$$y = x - 1$.

The equation of the tangent to the parabola $y = x^2 - x$ at the point where $x = 1$ is:

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