Find a possible formula for the quadratic function whose graph has vertex (2,6)and y intercept 9 y=?

Question

I need help learning this math bcuz it’s for college math and I have a test on it soon.

Answer 1

When you see the word parabola the standard equation that should be screaming at you is y=ax^2+ bx + c. In this problem, you are given the coordinates of the vertex (2, 6) and the y-intercept 9 which turns into the point (0, 9) 

To find the equation of the parabola we need to find the values of "a", "b" and "c" using the points given.

If you plug in the values of the y-intercept (given) into the standard equation...

y=ax^2 + bx + c

(9) = a(0)2 + b(0) + c.

c=9

Recalling the standard equation of the parabola and the vertices point given, plug in the vertices point as (x,y) and use the c value we just solved for to get an equation for b in terms of a. 

(2, 6) vertex point 

6= a(2)^2 + b(2) + 9 

-3 = 4a +2b

4a+2b+3=0

-4a-3=2b

B=-1/2a-3/2

Recall: c=9 

b=-1/2a-3/2

Now, let's solve system of equation: 

-3=4a+2b (from above)

b=-1/2a-3/2

Plug in for b

-3=4a+2(-1/2-3/2)

-3=-4a-1-3

0=-4a-1

1=4a

a=1/4

Solve for b

b=-1/2a-3/2 (from above)

b=-1/2(1/4)-3/2

-1/8-3/2=b

b=-1/8 - 12/8

b=-13/8

 

a = 1/4

b = -13/8

c = 9

y = 1/4x^2-13/8x+9