Question 103·200 Super-Hard SAT Math Questions·Advanced Math

The first quadrant is above the $x$-axis and right of the $y$-axis, so both coordinates must be positive: $x>0$ and $y>0$. A point on an axis does not count.

Type $y=x+1$ and $y=x^2-4x-5$ on separate Desmos lines. An intersection is on both graphs, so it solves both equations. Click the left crossing: Desmos shows $(-1,0)$. Its $x$-coordinate is negative, and it lies on an axis, so it is not in the first quadrant.

Click the crossing above and to the right of the axes. Desmos shows $(6,7)$, whose coordinates are both positive. Read the first coordinate for $x$, not the second. The requested value is $x=6$. Choice A.

At an intersection, the line and parabola have the same $y$-value. Set their expressions for $y$ equal:

To use factoring, get zero on one side. Subtract $x+1$ from both sides:

Look for two numbers that multiply to $-6$ and add to $-5$. They are $-6$ and $1$, so factor the expression:

The zero-product property says that if a product is zero, at least one factor is zero. Set each factor equal to zero: $x-6=0\quad\text{or}\quad x+1=0$. Solve each equation:

The negative root cannot be in the first quadrant. For $x=6$, use the line to check $y$: $y=6+1=7$. Both coordinates of $(6,7)$ are positive, so the requested $x$-value is $6$. Choice A.