Find b and c so that y=9x^2+bx+c has a vertex of (4,9)

Answer 1

Hi Lauren A

b = -72

c = 153

You are given a parabola in Standard or General Form f(x) = y = ax² + bx + c and the coordinates of its vertex (4,9), h and k where h is the x coordinate of the vertex and k is the y coordinate.

Given

y = 9x² + bx + c

where a = 9

vertex = (4,9), h = x = 4 = -b/2a and k = y = 9

You have a, x and y for your parabola

You can use and follow the given standard format and the relationship between the variables to solve and check the problem.

The coordinates of the vertex, h = x = -b/2a substituting

4 = -b/2(9)

4 = -b/18

4(-18) = b

-72 = b

Since we have b and we know x and y from the vertex coordinates we can simply solve for c as follows:

9 = 9(4)² - 72(4) + c

9 = 9(16) - 288 + c

9 = 144 -288 + c

9 = -144 + c

Add 144 to both sides of the equation to solve for c

9 + 144 = c

153 = c

Now that you have the variables, your standard form equation is

y = 9x² - 72x +153

In vertex form

y = 9(x - 4)² + 9

You can graph both equations on a graphing calculator if you have one or at Desmos.com to check and confirm the coordinates of the vertex.

Answer 2

9x^2 + bx + c

= 9(x^2 +bx/9) + c

= 9(x^2 +bx/9 + b^2/18^2) + c - 9b^2/18^2

= 9(x+b/18)^2 + c -b^2/36

b/18 = -4

b = -4(18) = -72

c-b^2/36 = 9

c -(-72)^2/36 = 9

c = 9 -72^2/36 = 9-144 = -135

b= -72, c= -135

y= 9x^2 -72x -135

y = 9(x^2 -8x) -135

y = 9(x^2-8x +16) -135 + 16(9)

y = 9(x-4)^2 -135 +144

y = 9(x-4)^2 + 9 in vertex form

y = 9(x-h)^2 +k where (h,k) = vertex = (4,9)