Parabolic Graph

Example 3  Graph  .

Solution
This is a parabola in the general form.
In this form, the x-coordinate of the vertex (the highest or lowest point on the parabola) is  and we get the y-coordinate is .  So, for our parabola the coordinates of the vertex will be.
                  
So, the vertex for this parabola is (1,4).

We can also determine which direction the parabola opens from the sign of a.  If a is positive the parabola opens up and if a is negative the parabola opens down.  In our case the parabola opens down.

Now, because the vertex is above the x-axis and the parabola opens down we know that we’ll have x-intercepts (i.e. values of x for which we’ll have  ) on this graph.  So, we’ll solve the following.
                                                             
So, we will have x-intercepts at  and .  Notice that to make our life easier in the solution process we multiplied everything by -1 to get the coefficient of the  positive.  This made the factoring easier.

Here’s a sketch of this parabola.

*ART :- ( Starters Focus only on equation of 2 degree is parabolic)
  1. First check the standard equation containing 3 constants a,b,c.
  2. If a>0 than graph will be upfacing (Case of Minima or Concave up ). If we know calculus Double derivative will be always greater than zero
  3. In order to find the minimum Point Equate single Derivative to Zero.
  4. If a<0 than graph is downfacing (i.e current example ). Case of Maxima and shape is Convex up.
  5. In that case Maximum Point can be fount after equating single derivative to zero. 
  6. On putting x=0 y comes (0,3) Thus graph will be shifted up from Origin. 
  7. Take Roots and plot them.
PS : Single derivative - Differentiate the above equation once
        Double Derivative Differentiate the equation twice ( consecutively )


CommonGraphs_G3

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