Finding the vertex of a parabola when knowing 2 points and the x of the vertex

Question

The parabola with the equation y=a(x-2)^2 +q. Passes through (-2,5) and (4,-1)  how do i solve for q? Thanks

Answer 1

y = a(x - h)² + k

 

By inspection,

 

h = 2

k = q

 

Use the 2 given points to solve 2 equations in 2 unknowns a and q.

 

5 = a(-2 - 2)² + q ⇒ 5 = 16a + q

-1 = a(4 - 2)² + q ⇒ -1 = 4a + q

 

Solve for a and q from:

 

5 = 16a + q

-1 = 4a + q

 

6 = 12a ⇒ a = 1/2

 

5 = 16(1/2) + q = 8 + q ⇒ q = -3

 

The vertex is at (h,k) = (2,-3)

 

The equation of the parabola is

 

y = (1/2)(x - 2)² - 3

 

Test the answers against the 2 given points.

Answer 2

Since (-2,5) is on the graph, we have 5 = a(-2-2)² + q.  So, 16a + q = 5

 

Since (4,-1) is on the graph, we get -1 = a(4-2)² + q.  So, 4a + q = -1

 

We have the system of equations:  16a + q = 5

                                                     4a + q = -1

 

Subtracting the equations, we obtain 12a = 6

 

                                                         a = 1/2

 

4a + q = -1 and a = 1/2, so 2 + q = -1  

 

                                               q = -3