Step-by-step explanation:
the standard form for a quadratic function is
y = ax² + bx + c
our parabola had the equation
y = a(x - 2)² + q
let's do the "math" :
y = a(x² - 4x + 4) + q = ax² - 4ax + 4a + q
so, we know (when comparing with the standard form)
b = -4a
c = 4a + q
for the vertex we need to find the axis of symmetry first by using the standard formula
x = -b/2a
which is in our case
x = -(-4a)/2a = 4a/2a = 2
to get y for this x we need to find a and q first.
we are using the given points for that :
5 = a(-2)² - 4a×(-2) + 4a + q
-1 = a(4)² - 4a×4 + 4a + q
the first equation then forms into
5 = 4a + 8a + 4a + q = 16a + q
the second equation forms into
-1 = 16a - 16a + 4a + q = 4a + q
let's subtract the second from the first equation
5 = 16a + q
- -1 = 4a + q
--------------------
6 = 12a + 0
12a = 6
a = 1/2
-1 = 4a + q = 4×1/2 + q = 2 + q
q = -3
so, the original equation for our parabola is
y = 1/2 × (x - 2)² - 3
so y for the vertex (remember, x = 2) is
y = 1/2 × (2 - 2)² - 3 = 1/2 × 0 - 3 = -3
so the vertex is (2, -3)
