### Instruction: Determine if the graph of the equation below is a parabola, circle, ellipse, hyperbola, point, line, two lines, or empty.

$x^2 + 2y^2 - 6x - 8y + 21 = 0$.### Response: To identify the type of graph, I will try to complete the square for both x and y terms and see what form I get.
First, I will group the x and y terms together and move the constant to the other side:
$(x^2 - 6x) + (2y^2 - 8y) = -21$
Then, I will factor out the coefficients of the squared terms and add the missing squares to both sides:
$(x^2 - 6x + 9) + 2(y^2 - 4y + 4) = -21 + 9 + 8$
I can simplify this by writing the perfect squares as binomials and combining the constants:
$(x - 3)^2 + 2(y - 2)^2 = -4$
This looks like the equation of an ellipse, except that the right-hand side is negative, which is impossible for an ellipse.
In fact, this equation has no real solutions, because the left-hand side is always positive or zero, and can never equal a negative number.
Therefore, the graph of this equation is empty, meaning it has no points on the coordinate plane.