Find a function whose graph is a parabola with this vertex and passes through this point

Question

Parabola with vertex (7,8) that passes through the point (6,-9)

Answer 1

Billy,

 

If the parabola is in form y=a(x-b)^2+c=0, then the vertex occurs at (b,c).  How do we know this?  If a >0 than y>c and the vertex is a minimum.  If If a<0 than y<c and the vertex is a maximum.

 

So y=a(x-7)^2+8 represents a family of parabolas with vertex at (7,8)

 

We can solve for a by requiring the parabola pass through the point (6,-9).  Substitute (6,-9) for (x,y).

 

-9=a(6-7)^2+8  So a=-17

 

y=-17(x-7)^2+8  You can convert to other forms as required.

 

 

 

 

Answer 2

Parabola with vertex (7,8) that passes through the point (6,-9). What follows is a systematic way of solving this.

 

The equation for a parabola is y = a x^2 + b x + c and we need to find the constants a, b and c.

 

The parabola passes through the vertex at (7,8) so substitute x = 7 and y = 8.

         8 = a 7^2 + b 7 + c

         8 = 49 a + 7 b + c

 

At the vertex the slope is 0. The slope is the derivative of the equation.

  The derivative of y = a x^2 + b x + c  is:

        dy/dx = a 2x + b

  The slope at the vertex is 0 so the derivative dy/dx = 0 and x = 7

         0 = a 2(7) + b

         0 = 14 a + b

 

Also the parabola passes through (6,-9) so substituting x = 6 and y = -9

        -9 = a 6^2 + b 6 + c

        -9 = 36 a + 6 b  + c

 

There are 3 equations in 3 unknowns. This is a very standard problem. We need to solve these systematically.

   (e1)  8 = 49 a + 7 b + c

   (e2)  0 = 14 a + b

   (e3) -9 = 36 a  + 6 b + c

 

Starting form the bottom. Lets eliminate the "c" in (e3). Subtract (e1) from (e3). First multiply (e1) by -1

   (-1 e1)  -8 = -49 a - 7 b  - c

   (e3)      -9 =  36 a + 6 b + c

    now adding term by term

   (e3')   -17 = -13 a - b

 

Now we have:

   (e1)     8 = 49 a + 7 b + c
   (e2)     0 = 14 a + b
   (e3') -17 = -13 a - b

 

Next eliminate the "b" from (e3') by adding (e2) to (e3')

   (e2)     0 = 14 a + b

   (e3') -17 = -13 a - b

   adding term by term

   (e3'')   -17 =  a

   This gives us a = -17

 

Substituting a = -17 into (e2)

    (e2) 0 = 14 (-17) + b

           0 = -238  + b

           b = 238

 

Substituting a and b into (e1)

     (e1) 8 = 49 a + 7 b + c

     (e1) 8 = 49 (-17) + 7 (238) + c

            8 = - 833 +1666 + c

            8 = 833 +  c

            8 - 833 = c

            c = - 825

 

Finally   y = -17 x^2 + 238 x - 825

   Checking the slope at x = 7

   dy/dx = 2 (-17) x + 238 

            = 2 (-17) 7 + 238

            = -238 + 238

            = 0  YES.

 

   Checking that y = 8 when x = 7

            y = -17 x^2 + 238 x - 825

            y = -17 (7)^2 + 238 (7) - 825

            y = -833 + 1666 - 825

            y = 8  YES.

 

    Checking that y = -9 when x = 6

           y = -17 x^2 + 238 x - 825

           y = -17 (6^2) + 238 (6) - 825

           y =   -612        +1428    - 825

           y = -9   YES.