At which point on the curve y = x^2 + 3x shou | vedclass

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(A) Let the point on the curve be $P(x_1, y_1)$. Since $P$ lies on the curve $y = x^2 + 3x$,we have $y_1 = x_1^2 + 3x_1$. The slope of the tangent at

Vedclass · Accepted Answer

$(-3, 0)$

Vedclass · Answer

(A) Let the point on the curve be $P(x_1, y_1)$. Since $P$ lies on the curve $y = x^2 + 3x$,we have $y_1 = x_1^2 + 3x_1$. The slope of the tangent at $(x_1, y_1)$ is given by $\frac{dy}{dx} = 2x + 3$. So,the slope $m = 2x_1 + 3$. The equation of the tangent at $(x_1, y_1)$ is $y - y_1 = (2x_1 + 3)(x - x_1)$. Since the tangent passes through $(0, -9)$,we substitute $x = 0$ and $y = -9$: $-9 - y_1 = (2x_1 + 3)(0 - x_1)$ $-9 - y_1 = -2x_1^2 - 3x_1$ $y_1 = 2x_1^2 + 3x_1 - 9$. Equating the two expressions for $y_1$: $x_1^2 + 3x_1 = 2x_1^2 + 3x_1 - 9$ $x_1^2 = 9$ $x_1 = \pm 3$. If $x_1 = 3$,then $y_1 = (3)^2 + 3(3) = 18$. Point is $(3, 18)$. If $x_1 = -3$,then $y_1 = (-3)^2 + 3(-3) = 0$. Point is $(-3, 0)$. Comparing with the given options,the correct point is $(-3, 0)$.

At which point on the curve $y = x^2 + 3x$ should the tangent be drawn so that it passes through the point $(0, -9)$?

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