Find the quadratic function

Question

Find the quadratic function y=ax2 + bx + c whose graph passes through the given points. (0, -5), (-2, -7), (1, 14)

Answer 1

y= ax^2 +bx +c Replace the first giving point (0, -5) in the function y=-5 and x=0

1) . -5 = a(0)^2 +b(0) +c . c=-5 and I got c=-5

y=ax^2 +bx +c

2) . -7 =a(-2)^2+b(-2) -5 Replacing the second giving point . (-2, -7)

-7+5 = 4a-2b

4a -2b = -2 . I just create Equation A

Replacing (1, 14)

y= ax^2 +bx +c

14=a(1)^2 +b(1) -5

14+5 = a+b

19= a+b . Equation B

Now lets combine equation A and equation B

4a - 2b =-2

a +b =19

___________

Using the substitution method to find a and b. from equation B a= 19-b . let substitute a in equation A

4(19-b) -2b =-2

76-4b-2b=-2

-6b =-78

b=13 . and a= 19-b a= 19-13=6

so the equation or function that you are looking for is . y = ax^2 +bx +c . or y=6x^2+13x-5

Note: In order to check your answer graph the function in a ti-84, or desmos graphing calculator, then press 2nd GRAPH and notice the points (-2,-7),(-1,-12), (0,-5) and (1,14) these points were giving to you where the function passing trough.

If you need extra help. just let me know. Good luck during the new school year coming up!!

Answer 2

Quadratic; x=-b±√(b^2-4ac)(2a)