The expression y = ax^2 + bx + c always has t | vedclass

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Question

(B) Let $f(x) = ax^2 + bx + c$. Then $f(0) = c$. Thus,the graph of $y = f(x)$ meets the $y$-axis at $(0, c)$.If $c > 0$,then by hypothesis $f(x) > 0$.

Vedclass · Accepted Answer

$4ac > b^2$

Vedclass · Answer

(B) Let $f(x) = ax^2 + bx + c$. Then $f(0) = c$. Thus,the graph of $y = f(x)$ meets the $y$-axis at $(0, c)$.If $c > 0$,then by hypothesis $f(x) > 0$. This means that the curve $y = f(x)$ does not meet the $x$-axis.If $c Thus,in both cases,$y = f(x)$ does not intersect the $x$-axis,i.e.,$f(x) eq 0$ for any real $x$.Hence,$f(x) = 0$,i.e.,$ax^2 + bx + c = 0$ has imaginary roots,which implies the discriminant $D = b^2 - 4ac < 0$,or $b^2 < 4ac$.

The expression $y = ax^2 + bx + c$ always has the same sign as $c$ if

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