Let the parabola y = x^2 + px + q passing thr | vedclass

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(B) The parabola is given by $y = x^2 + px + q$. Since it passes through $(1, -1)$, we have $-1 = 1 + p + q$, which implies $q = -p - 2$.The vertex of

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$4$

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(B) The parabola is given by $y = x^2 + px + q$. Since it passes through $(1, -1)$, we have $-1 = 1 + p + q$, which implies $q = -p - 2$.The vertex of the parabola $y = ax^2 + bx + c$ is at $(-b/2a, -D/4a)$. For $y = x^2 + px + q$, the vertex is $(-p/2, q - p^2/4)$.The distance from the vertex to the $x$-axis is the absolute value of the $y$-coordinate of the vertex, which is $d = |q - p^2/4|$.Substituting $q = -p - 2$, we get $d = |-p - 2 - p^2/4| = |p^2/4 + p + 2|$.To minimize $d$, we analyze the quadratic $f(p) = p^2/4 + p + 2$. The derivative $f'(p) = p/2 + 1$. Setting $f'(p) = 0$ gives $p = -2$.At $p = -2$, $q = -(-2) - 2 = 0$.The value of $p^2 + q^2 = (-2)^2 + 0^2 = 4 + 0 = 4$.

Let the parabola $y = x^2 + px + q$ passing through the point $(1, -1)$ be such that the distance between its vertex and the $x$-axis is minimum. Then the value of $p^2 + q^2$ is:

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