Find in the form y= ax^2 bx c, the equation of the quadratic whose graph a) touches the xaxis at 4 and passes through (2,12) b) has vertex (4,1) and passes through (1,11)Show that the vertex of the parabola ax^2 bx c is at (b/2a, 4acb^2/4a) I'm guessing this has something to do with the roots being at (b√(b^24ac))/2a but I'm not sure what I do after that Answer Save 6 Answers Relevance Old Teacher Lv 7 10 years ago Favourite answerA quadratic function in the form y = ax 2 bx c is not always simple to graph We do not know the vertex or the axis of symmetry simply by looking at the equation To make the function easier to graph, we need to convert it to the form y = a(x h) 2 k
A Find The Vertex Form Equations Y A X H 2 K For The Parabolas Below Assume Parameter A Brainly Com
