Question

Determine the vertex and the axis of symmetry for the function below:

y = -x^2 + 2x + 1
a.
vertex: (1, 2)
axis of symmetry: y = 1
c.
vertex: (2, 1)
axis of symmetry: y = 1
b.
vertex: (-1, 2)
axis of symmetry: x = 1
d.
vertex: (1, 2)
axis of symmetry: x = 1

Answer 1

for \(f(x) = ax^2+bx+c\)
\(\text{vertex} = (-\frac{b}{2a},f(-\frac{b}{2a}))\)
\(\text{axis of symmetry }\rightarrow x = -\frac{b}{2a}\)