How do you graph 2x^2+x-14?

Question

I know that the vertex is (-0.3, -14.1) but I need a step-by-step on how to graph it. Please help!

Answer 1

 Y = 2 X^2 + X  -14

 

    a is positive, therefore parabola opens upward

 

      X = - 1/ 4 ± √[(1^2 - 4 ( 2) ( -14) ] / 4

 

          = -1/4 ±√( 1^2 + 112) /4

 

           = - 1 ±√113

                   4

     Axis of Symmetry :   X = - 1/4

 

       This has irrational root , and can not locate precisely the X-intercepts :

 

            Vertex ( -1/4 , -113/8)

 

               Draw axis of Symmetry :  X = -1/4

 

                  Locate the Vertex on the graph

 

                   Connect the vertex to Y - intercept( 0 - 14), this is one side of theparabola

                   Measure the same distance on X intercept, and mark , then connect vertex to the marked

                   point is the rough sketch of the graph.

 

 

 

Answer 2

You got the vertex already, so, the hard part is done.  You notice that the coefficient to the x^2 term is positive and greater than 1.  This means the parabola will open up and be narrow.  The next step would be to find x-values to the left and right of the vertex (preferably 2 or more each).  When you determine the values you want to use, plug them into the equation to get your y-values.  Now, take your ordered pairs and plot them in the graph.  Finally, connect the points.

Answer 3

Hi La;
Unfortunately, I do not have such a graphing mechanism available through Wyzant.

I should also have established the y-intercept for you.

y-intercept, value of y when x=0...

y=2x^2+x-14

y=-14

(0,-14)

 

Otherwise, plot a few numbers...

x=-1, y=-13

x=1, y=-11

and draw the parabola.