classify the graph of the equation -4y^2+5X+3y+7=0 as a circle, a parabola, an ellipse, or a hyperbola.
since there's y^2 but no x^2, it's a parabola, opening to the right:
5x = 4y^2 - 3y - 7
x = 4/5 y^2 - 3/5 y - 7/5
x = 4/5 (y^2 - 3/4 y) - 7/5
x = 4/5 (y^2 - 3/4 y + (3/8)^2) - 7/5 - 4/5 * (3/8)^2
x = 4/5 (y - 3/8)^2 - 121/80
so the vertex is at (-121/80,3/8)
http://www.wolframalpha.com/input/?i=plot+-4y^2%2B5X%2B3y%2B7+%3D+0
To classify the graph of the equation -4y^2 + 5X + 3y + 7 = 0, we need to examine the equation and determine its standard form for different conic sections.
1. Circle: General form is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle and r is the radius. The given equation does not fit the general form of a circle, so it is not a circle.
2. Parabola: General form is y = ax^2 + bx + c or x = ay^2 + by + c. Since the given equation is not in either of these forms, it is not a parabola.
3. Ellipse: General form is (x - h)^2 / a^2 + (y - k)^2 / b^2 = 1, where (h, k) is the center of the ellipse and a and b represent the semi-major and semi-minor axes respectively. The given equation is not in the general form of an ellipse, so it is not an ellipse.
4. Hyperbola: General form is (x - h)^2 / a^2 - (y - k)^2 / b^2 = 1 or (y - k)^2 / b^2 - (x - h)^2 / a^2 = 1, where (h, k) is the center of the hyperbola and a and b represent the distance from the center to the vertices, along the transverse and conjugate axes. The given equation does not fit the general form of a hyperbola, so it is not a hyperbola.
Hence, based on the analysis above, the given equation does not represent a specific conic section.