Given y=x^2+bx+c the vertex is -4.-7. Find the missing coefficients

Question

I was never told how to do this by my teachers.....

Answer 1

Hi Teegan

This is a problem where you want to be able to make use of the Vertex Form in order to get the values of b and c for the Standard Form of the parabola.

vertex (-4, -7) or (h, k) in the standard form these are the x and y coordinates of the vertex

Note the Standard form is

ax² + bx + c

Since you are given

x² + bx + c

a = 1

Note the vertex form is

y = a(x - h)² + k

Since a = 1

y = (x -(-4))² - 7

y = (x + 4)² - 7

y = x² + 8x + 16 - 7

y = x² + 8x + 9

You can check the vertex in the Standard Form as follows:

x² + bx + c

a = 1

You can use x coordinate of the vertex = -b/2a to confirm b

-4 = -b/2(1)

-4 = -b/2

2(-4) = -b

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

8 = b

You can also confirm the y coordinate of the vertex in the standard form by plugging in the x coordinate

y = (-4)² + 8(-4) + 9

y = 16 - 32 + 9

y = -16 + 9

y = -7

You can plot both at Desmos.com to see that they are the same.

Answer 2

Since you are given the vertex you can write the quadratic in vertex form y=a(x-h)^2 +k where the coordinates of the vertex are (h,k) and a would be the coefficient of the x^2 term

So you have y= (x+4)^2 -7

Y= x^2 +8x+16-7

Y= x^2 +8x+9

So b=8 and c=9