A free website which can solve this equationsolve for v, h ( v − t ) = h ( v + t ) where h ( x ) = a x 2 + b x + c. The value is the vertex of the quadratic function the website wolfarmalpha can solve that when a,b and c are numbers. I want the students to figure the vertex of the quadratic function.

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A free website which can solve this equationsolve for v,   h  (  v  −  t  )  =  h  (  v  +  t  ) where   h  (  x  )  =  a      x    2    +  b  x  +  c. The value is the vertex of the quadratic function the website wolfarmalpha can solve that when a,b and c are numbers. I want the students to figure the vertex of the quadratic function.

apopihvj · Accepted Answer

Step 1You have  a  (  v  −  t      )    2    +  b  (  v  −  t  )  +  C  =  a  (  v  +  t      )    2    +  b  (  v  +  t  )  +  CStep 2It simplifies to  2  a  v  t  +  b  t  =  0Solve for v to get  v  =            −      b              2      a

bucstar11n0h · Accepted Answer

Step 1I think Geogebra can handle a particular case for your question on demonstrating the shifting effect.For the s/w to have something to plot, we give it a sample curve like   h  (  x  )  =      x    2    −  x  −  12.Step 2Next, we ask the s/w to plot   y  =  h  (  x  −  2  ), where 2 is just particular value for t. The result clearly shows the resultant curve is just a right shift of the original by 2 units.Plotting   y  =  h  (  x  +  2  ) is a left shift of the original.Solving   h  (  x  −  t  )  =  h  (  x  +  t  ) is to find the point of intersection of the plotted curves.Since the original quadratic is symmetric (with   x  =            −      b              2      a       being the axis), the two plotted curves are just displaced right and left of the same amount. They will meet at   x  =            −      b              2      a

A free website which can solve this equation. solve for v , h(v-t)=h(v+t) where h(x)=ax^2+bx+c.

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