What are the lengths of the intercepts on the | vedclass

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(C) Given the curve $y = 2x^2 + 3x - 2$. First,find the derivative: $\frac{dy}{dx} = 4x + 3$. At the point $(1, 3)$,the slope of the tangent is $m = 4

Vedclass · Accepted Answer

$4/7, -4$

Vedclass · Answer

(C) Given the curve $y = 2x^2 + 3x - 2$. First,find the derivative: $\frac{dy}{dx} = 4x + 3$. At the point $(1, 3)$,the slope of the tangent is $m = 4(1) + 3 = 7$. The equation of the tangent line at $(1, 3)$ is given by $y - 3 = 7(x - 1)$,which simplifies to $y - 3 = 7x - 7$,or $7x - y = 4$. To find the $x$-intercept,set $y = 0$: $7x = 4 \implies x = 4/7$. To find the $y$-intercept,set $x = 0$: $-y = 4 \implies y = -4$. Thus,the intercepts on the coordinate axes are $4/7$ and $-4$.

What are the lengths of the intercepts on the coordinate axes made by the tangent to the curve $y = 2x^2 + 3x - 2$ at the point $(1, 3)$?

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