parabola of the form y=ax2+bx+c has a maximum value of y=4. The y-cord the parabola at x=4 is -4/3. The y-coordinate of the parabola at x=6 is -52/3.

Question

smaller and larger values of
determine the x intercepts

Answer 1

Plug in the values of x and y for each case in the general form of the parabola.

 

For the point (4, -4/3):

 

a(4)² + b(4) + c = -4/3

16a + 4b + c = -4/3

48a + 12b + 3c = -4                        eq1

 

 

For the point (6, -52/3):

 

a(6)² + b(6) + c = -52/3

36a + 6b + c = -52/3

108a + 18b + 3c = -52                   eq2

 

 

For the maximum point:        x = -b/(2a)

 

a(-b/2a)² + b(-b/2a) + c = 4

 

(b² / 4a²)a - (b² / 2a) + c = 4

 

(b² / 4a) - (b² / 2a) + c = 4

 

 

Multiplying all sides of the equation by 4a,

 

b² - 2b² + 4ac = 16a

 

-b² + 4ac = 16a

 

 

Using algebraic manipulation,

 

-b² = 16a - 4ac

 

-b² = a(16 - 4c)

 

b² = 4ac - 16a

 

b = √(4a - 16a)                        eq3

 

 

The bolded equations will be your system of equations. 

 

Substitute the value of b from eq3 into eq1 and eq2.  Now eq1 and eq2 will be in terms of a and c.  Then you can use elimination/substitution method to solve for a and c using those 2 equations.  Then once you have the values of a and c, substitute them into eq3 to solve for b.

 

Note:  Since you have a maximum, the value of variable "a" you obtain must be negative.

 

Once you have a, b, and c together, plug them into the parabola formula to get your actual formula.

 

Lastly, use the quadratic formula to find your x-intercepts.

 

 

I have guided you through all the hard stuff.  It is up to you to finish up here.