The x and y-intercepts of a quadratic function in the form

Question

y=a(x-h)^2+k

 

are (2,0), (-4,0), (0,16).

 

Solve for a, h, and k.

 

Present answer in form y=a(x-h)^2+k.

Answer 1

Because of (2,0) is a x-intersect of function y, we have:

 

0 = a(2-h)^2 + k  (1)

 

Because of (-4,0) is a x-intersect of function y, we have:

 

0 = a(-4-h)^2 + k = a(4+h)^2 + k  (2)

 

Because (0,16) is a y-intersect of function y, we have:

 

16 = a(0-h)^2 + k =  ah^2 + k  (3)

 

We have to find the values of h, k and a which are solutions of the following equation system:

 

a(2-h)^2 + k = 0     (1)

a(4+h)^2 + k = 0    (2)

ah^2 + k = 16         (3)   

 

From eq. (2) we have:  a(4+h)^2+k = 0

                                  a(16 + 8h +h^2) + k = 0

                                  16a + 8ah + ah^2 + k = 0

                                  16a + 8ah +(ah^2 +k) = 0  (2.1)

 

Substituting eq. (3) into eq. (2.1) we have:     16a + 8ah + 16 = 0

                                                                     2a  + ah  +  2  = 0  

                                                                     2a  + ah = -2    eq. (2.2)

                                                                    

From eq. (1) we have:  a(2-h)^2 + k = 0

                                  a(4 -4h +h^2) + k = 0

                                  4a - 4ah + ah^2 + k = 0

                                  4a -4ah + (ah^2 +k) = 0    (1.1)

 

Substituting eq. (3) into eq. (1.1) we have:   4a - 4ah + 16 = 0

                                                                 4a  - 4ah  = -16

                                                                   a - ah = -4      eq. (1.2)

 

Finally, we have only two equations (1.2) and (2.2) to be solved:

 

2a + ah = -2

 a  - ah = -4    Using the procedure of elimination:

 

 (x1)      2a + ah = -2

 (x1)        a  - ah  = -4      

--------------------------------

             3a      0    = -6               a = -2

 

Substituting this value of "a" into eq. (1.2), we have:   a - ah = -4   eq. (1.2)

                                                                               -2 -(-2)h = -4

                                                                               -2 +2h = -4

                                                                                    2h = -2        h = -1

 

Substituting these values of "a" and "h" into eq. (3), we have:   ah^2 + k = 16    eq. (3)

                                                                                            -2(-1)^2 + k = 16

                                                                                            -2 + k = 16                   k= 18

 

Answer 2

Point (2, 0): 0 = a(2-h)² + k

Point (-4, 0): 0 = a(-4-h)² + k

Point (0, 16): 16 = a(-h)2 + k = ah² + k

 

From the first two equations:

0 = a(2-h)² + k = a(-4-h)² + k

a(2 - h)² = a(-4-h)²

(2-h)² = (-4-h)²

h²- 4h + 4 = h² + 8h + 16

12h = - 12

h = -1

 

From the third equation, with h = -1:

16 = a + k

k = 16 - a

 

Substitute this value for k and h = -1 into the first equation:

0 = 9a + 16 -a

8a = -16

a = -2

 

Then k = 16 - a = 18

 

Since the values of a, h, and k are known the equation is

y = -2(x+1)² + 18