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Q.
The tangent to the curve $y = x^2 + 3x$ will pass through the point $(0, -9)$ if it is drawn at the point
Application of Derivatives
Solution:
Equation of tangent at the point $(x, y)$ is
$Y - y = (2x + 3)(X-x)$
If it passes through $(0, -9)$, then
$- 9 - y = (2x + 3)(0 - x)$
$\Rightarrow $ $y + 9 = 2x^2 + 3x$
$\Rightarrow $ $x^2 + 3x + 9 = 2x^2 + 3x$
$\Rightarrow $ $x^2 = 9$
$\Rightarrow $ $x = ± 3$
$\therefore $ $y = 18$ or $0$.
$\therefore $ $(-3,0)$ is the choice out of given points.