AMC 12 Mock Exam 2 Problem 6

Problem 6 of 25

The graph of $y=x^2+bx+c$ crosses the $x$ -axis at two points that are $6$ units apart, and the vertex of the graph lies on the line $y=3x+6$ . What is $c$ ?

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Solution

Write $y=(x-h)^2+k$ with vertex $(h,k)$ . The roots are $h\pm\sqrt{-k}$ , which are $2\sqrt{-k}=6$ apart, so $k=-9$ . The vertex lies on $y=3x+6$ , so $-9=3h+6$ and $h=-5$ . Then $c$ is the value at $x=0$ : $c=h^2+k=25-9=\boxed{16}$ .

Tagged: quadratics: vertex form and root spacing

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