Graph the parabola {eq}y = x^2 {/eq} Find the following 1) a and h 2) Vertex 3) Equation of axis of symmetry 4) Vertex maximum or minimum 5) Transformation from standard {eq}y = x^2 {/eq}Find the vertex and focus of y2 6y 12x – 15 = 0 The y part is squared, so this is a sideways parabola I'll get the y stuff by itself on one side of the equation, and then complete the square to convert this to conics form y2 6 y – 15 = –12 x y2 6 yIf the graph is a parabola, find its vertex If the graph is a circle, find its center and radiusx2 y2 6x 6y 2 = 0 asked in Mathematics by Brooke

Using Properties Of Parabolas To Graph A Parabola Algebra And Geometry Help