### Instruction: How many $y$-intercepts does the graph of the parabola $x = 2y^2 - 3y + 7$ have?### Response: I know that the $y$-intercept of a graph is the point where it crosses the $y$-axis, or where $x = 0$.
So, to find the $y$-intercepts of the parabola, I need to solve the equation $0 = 2y^2 - 3y + 7$.
This looks like a quadratic equation, so I could try to factor it, use the quadratic formula, or complete the square.
I don't see any obvious factors, so I'll skip that method.
To use the quadratic formula, I need to identify the coefficients $a, b, c$ of the equation in the form $ax^2 + bx + c = 0$.
In this case, I have $a = 2, b = -3, c = 7$.
The quadratic formula says that the solutions are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Plugging in the values, I get $y = \frac{3 \pm \sqrt{9 - 56}}{4}$.
Simplifying, I get $y = \frac{3 \pm \sqrt{-47}}{4}$.
Since the square root of a negative number is not a real number, this means that there are no real solutions for $y$.
Therefore, the graph of the parabola does not cross the $y$-axis at any point, and there are no $y$-intercepts.