The parabolic graph of a quadratic equation has a minimum at (-6,-18) and passes through the point (0,54). Find this quadratic equation.

Question

Need some help with this problem. Thanks in advance.

Answer 1

Hi Anthony W.

A parabola associated with the quadratic equations

f(x) = ax² + bx + c

f(x) = a(x - h)² + k

Has only one minimum or maximum and opens up or down; the coordinates (-6,-18), the minimum are the lowest point on your parabola then it opens up from there. Also you are given the point where your parabola crosses the y axis; (0,54) this is the y intercept. The y intercept given also indicates that your parabola opens upward. You can use these points in the appropriate equation above to determine the equation of your parabola.

If you look up these equations and parabolas in general you will find there is a special name associated with maximum or minimum point, it is called the vertex with coordinates (h,k).

Since you have (h,k) and the y intercept you can solve for the missing part to determine your quadratic equation.

You can check your equation by graphing it at Desmos.com to confirm that it passes through all the points given.

I got f(x) = 2x² + 24x + 54

Can you look up the information and check my result using the other equation above, just give a try.

Answer 2

Since the parabolic graph has a minimum as its vertex, therefore it is open up. The vertex form is:

y = a (x - h)² + k, and we expect variable a to be positive.

V(h,k) = (-6,-18)

(x,y) = (0,54)

54 = a (0 + 6)² - 18

54 = 36a - 18

-36a = -54 - 18

-36a = -72

a = 2

Therefore our quadratic equation is:

y = 2(x + 6)² - 18