Suppose a parabola y=ax^2+bx+c has two x-inte | vedclass

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(B) The equation of the parabola is $y=ax^2+bx+c$. Since the vertex is $(2,-2)$ and the parabola opens upwards (as it has two $x$-intercepts and the v

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$bc > 0$

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(B) The equation of the parabola is $y=ax^2+bx+c$. Since the vertex is $(2,-2)$ and the parabola opens upwards (as it has two $x$-intercepts and the vertex is below the $x$-axis),we have $a > 0$. The $x$-coordinate of the vertex is given by $-\frac{b}{2a} = 2$. Since $a > 0$,we have $-b = 4a$,which implies $b = -4a$. Since $a > 0$,it follows that $b The $y$-intercept is at $x=0$,which is $y=c$. From the graph,the $y$-intercept is below the $x$-axis,so $c Now,consider the product $bc$. Since $b  0$. Thus,option $(b)$ is correct.

Suppose a parabola $y=ax^2+bx+c$ has two $x$-intercepts,one positive and one negative,and its vertex is $(2,-2)$. Then,which of the following is true?

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