The equation of the tangent to the curve y=5x | vedclass

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(C) Given the curve equation: $y=5x^2-3x+7$. First,we find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent: $\frac{dy}{dx} = \fra

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$13x+y+9=0$

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(C) Given the curve equation: $y=5x^2-3x+7$. First,we find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent: $\frac{dy}{dx} = \frac{d}{dx}(5x^2-3x+7) = 10x-3$. The slope of the tangent $m_T$ at the point $(-1, 4)$ is: $m_T = \left. \frac{dy}{dx} \right|_{x=-1} = 10(-1)-3 = -10-3 = -13$. Now,using the point-slope form of the line equation $y-y_1 = m(x-x_1)$: $y-4 = -13(x-(-1))$ $y-4 = -13(x+1)$ $y-4 = -13x-13$ $13x+y-4+13 = 0$ $13x+y+9 = 0$.

The equation of the tangent to the curve $y=5x^2-3x+7$ at the point $(-1, 4)$ is:

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