The graph of y=f(x) has the line x=5 as an axis of symmetry. The graph also passes through the point (-8, -7) Find another point that must lie on the graph of y=f(x).
+128408 Assuming a monic quadratic polynomial ....we must have the point (5, k) as the vertex
Since (-8, -7) is on the graph, we have that
-7 = ( -8 - 5)^2 + k
-7 = (-13)^2 + k
-7 = 169 + k
k = -7 -169 = -176
So the function is
y= (x - 5)^2 - 176
So......the point (5, -176) also must be on the graph
