A parabola passes through the point of intersection of the lines with equations $y=-x+3$ and $x-2 y-6=0$, as well as the the $x$-intercept of each line. If the parabola also passes through the point $(10, k)$, what is the value of $k$ ?
We first find the point of intersection of the two lines.

To do this, rearrange the equation of the second line to $y=\frac{1}{2} x-3$.

Setting the $y$-values equal to each other, we obtain $-x+3=\frac{1}{2} x-3$ and multiplying through by 2 gives $-2 x+6=x-6$, so $3 x=12$, or $x=4$.

To find the $y$-coordinate of the point of intersection, we can substitute $x=4$ into the first equation to get $y=-4+3=-1$.

Therefore, the parabola passes through the point $(4,-1)$.

To find the $x$-intercept of the first line, we set $y=0$ to get $0=-x+3$ so $x=3$. The $x$-intercept of the first line is 3 .

To find the $x$-intercept of the second line, we set $y=0$ to get $x-2(0)-6=0$, so $x=6$. The
$x$-intercept of the second line is 6 .

The parabola passes though $(3,0)$ and $(6,0)$, so its equation takes the form $y=a(x-3)(x-6)$. To solve for $a$, we substitute $(x, y)=(4,-1)$ to get $-1=a(4-3)(4-6)=-2 a$ or $-1=-2 a$, so $a=\frac{1}{2}$.

The equation of the parabola is $y=\frac{1}{2}(x-3)(x-6)$.

Since the parabola passes through the point $(10, k)$, we can substitute to get $k=\frac{1}{2}(10-3)(10-$ $6)=\frac{28}{2}=14$. Therefore, $k=14$.

ANSWER: 14

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