Respuesta :
the complete question in the attached figure
we know that
the equation of a parabola is
y=a(x-h)²+k
where
(h,k) is the vertex --------> (h,k)--------> (-1,2)
so
y=a(x+1)²+2
point (2,20)
for x=2
y=20
20=a(2+1)²+2--------> 20=a*9+2--------> 9*a=18---------> a=2
the equation of a parabola is
y=a(x+1)²+2-------> y=2(x+1)²+2
therefore
the answer is the option
C) f(x) = 2(x + 1)2 + 2
we know that
the equation of a parabola is
y=a(x-h)²+k
where
(h,k) is the vertex --------> (h,k)--------> (-1,2)
so
y=a(x+1)²+2
point (2,20)
for x=2
y=20
20=a(2+1)²+2--------> 20=a*9+2--------> 9*a=18---------> a=2
the equation of a parabola is
y=a(x+1)²+2-------> y=2(x+1)²+2
therefore
the answer is the option
C) f(x) = 2(x + 1)2 + 2

Answer:
Step-by-step explanation:
Given that The minimum of the graph of a quadratic function is located at (-1,2)
This implies that parabola is open up.
Hence parabola would have equaiton of the form
[tex]y-2 = 4a(x+1)^2[/tex]
To find a:
We use the fact that the parabola passes through (2,20)
Substitute x=2 and y =20
[tex]20-2 = 4a(2+1)^2\\18 =36a\\a = 0.5[/tex]
Hence equation would be
[tex]y-2 = 2(x+1)^2[/tex]