Quadratic models: Parabola graph of ax^2+bx+c

Question

A parabola graph has the Vertex and Y-intercept is A(0,8) and x-intercepts of B(3,0) and C(-3,0)....

 

Question 1) The parabola has the equation ax^2+bx+c , Use "A(0.8)" to find the value of "b" and "c" from the equation ax^2+bx+c

 

Question 2) Use B(3,0) to find value of "a" which belongs to ax^2+bx+c

Answer 1

The vertex of a parabolic equation can be written as y=a(x-b)^2 + c where the vertex is (b,c). We know this because x=b, y=c. Since we know the vertex is (0,8), that makes b=0 here and c=8 or y=ax^2+8. So the value of "b" is 0 and "c"=8.

 

Since this parabola has x intercepts of +/-3, that means we can plug in (3,0) into y=ax^2+8 to find 0=9a+8, or a=-8/9.

Answer 2

1) To use A(0,8) is the same as writing 8 = a(0)² + b(0) + c  and 0 = -b / 2a

    yielding b = 0 and c = 8

 

2) Using B(3, 0)  ⇒    0 =  a(3)² + 0(3) + 8   ⇔  0 = 9a + 8  ⇔  a = -8/9

 

y = (-8/9) x² + 8