text: # Assignment 3:Locus of Vertices of Parabolas of the Form y = x2 + bx + 1
  
  From Jim WilsonÕs Assignment page:  Consider the locus of the vertices of the set of parabolas graphed from .  Show that the locus is the parabola .
  
  By using a graphing utility, it appears that the vertices of these parabolas do lie along another parabolic curve.
  
  We can support this conjecture by tracing the vertex of the parabola , using geometerÕs sketchpad.  The red parabola in the image below was created by tracing the vertex and appears to be .
  
  We can now explore the situation algebraically.
  
  The vertex of any parabola, , lies on the parabolaÕs line of symmetry, .  So each vertex for a parabola  has  as itÕs abscissa.  The ordinate of each vertex is
  .
  
  ## So for each value b, we have a vertex .  Thus, vertices of the parabolas trace out the curve .
  
  In general, the vertices of , have abscissa .  The ordinate of a given vertex is
  
  So for any quadratic function , the locus of the vertex as b varies will be given by .
  
  For calculus students:  What is the locus of the point on a parabola whose tangent line has slope one?  Is there an algebraic generalization?  A rough sketch of a solution is here.
metadata: {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}