29/2/16 The #x^2# is positive so the general graph shape is #uu# Consider the generalised form of #y=ax^2bxc# The #bx# part of the equation shifts the graph left or right You do not have any #bx# type of value in your equation So the graph is central about the yaxis The #c# part of the equation is of value 1 so it lifts the vertex up from y=0 to y=1The graphs y = 2x^3 4x 2 and y = x^3 2x 1 intersect at exactly 3 distinct points The slope of the line passing through two of these points > 12th > Maths > Application of Derivatives > Tangents and NormalsSimple and best practice solution for y=x^32x^21 equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so
![How Do You Solve The System By Graphing Y 2x 1 And Y 2x 2 Socratic How Do You Solve The System By Graphing Y 2x 1 And Y 2x 2 Socratic](https://useruploads.socratic.org/wmw5I6l8R1eTS2njvDGc_graph-1.jpg)
How Do You Solve The System By Graphing Y 2x 1 And Y 2x 2 Socratic