### Instruction: The line $x = k$ intersects the graph of the parabola $x = -2y^2 - 3y + 5$ at exactly one point.  What is $k$?### Response: To find the value of $k$ that makes the line $x = k$ intersect the parabola $x = -2y^2 - 3y + 5$ at exactly one point, I need to solve the equation $k = -2y^2 - 3y + 5$ for $y$ and see when there is only one solution.
This equation is a quadratic equation in $y$, so I can try to use the quadratic formula to find the solutions.
The quadratic formula says that if $ax^2 + bx + c = 0$, then $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
In this case, $y$ is the variable, and $a = -2$, $b = -3$, and $c = 5 - k$.
So, the solutions for $y$ are $y = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(-2)(5 - k)}}{2(-2)}$.
Simplifying, I get $y = \frac{3 \pm \sqrt{9 + 8(5 - k)}}{-4}$.
Now, I want to find the value of $k$ that makes this expression have only one solution for $y$.
This means that the expression under the square root, $9 + 8(5 - k)$, has to be zero, because if it is positive, there are two solutions, and if it is negative, there are no real solutions.
So, I set $9 + 8(5 - k) = 0$ and solve for $k$.
Subtracting $9$ from both sides, I get $8(5 - k) = -9$.
Dividing both sides by $8$, I get $5 - k = -\frac{9}{8}$.
Subtracting $5$ from both sides, I get $-k = -\frac{9}{8} - 5$.
Multiplying both sides by $-1$, I get $k = \frac{9}{8} + 5$.
Adding $5$ to $\frac{9}{8}$, I get $k = \frac{49}{8}$.