Identify the vertex and axis of symetry of the parabola. Identify points corresponding to P and Q. on the graph it has P at (-2,4) and a point at Q at (-1,7). my teacher says the answers were the vertex was (-1,3) the Axis of symmtry was X=-1 the P is (0,4) and Q is (-3,7).. how do you figure out each of those steps? im confused with how to do the whole problem
2 answers
The P and Q coordinates in your first statement do not agree with the P and Q in your last statement. I would need to see the graph to know what this problem is all about.
I will try to reverse engineer this mess.
(y-3) = k (x+1)^2
if the vertex is at (-1,3) and x=-1 is the axis
put in P at (-1,7) impossible because y = 3 when x = -1
so I will try the second pair of points
P at (0,4)
(4-3) = k (1)^2
so k = 1
so
y - 3 = (x+1)^2
try Q at (-3,7)
4 = (-2)^2
YES
so the parabola with axis of symmetry at x = -1 and vertex at (-1,3) can contain the SECOND pair of points P and Q
rewrite in standard form
y = 3 +x^2 + 2 x + 1
y = x^2 + 2 x + 4
I have no idea what the two original points P and Q were. They are on some other curve.
(y-3) = k (x+1)^2
if the vertex is at (-1,3) and x=-1 is the axis
put in P at (-1,7) impossible because y = 3 when x = -1
so I will try the second pair of points
P at (0,4)
(4-3) = k (1)^2
so k = 1
so
y - 3 = (x+1)^2
try Q at (-3,7)
4 = (-2)^2
YES
so the parabola with axis of symmetry at x = -1 and vertex at (-1,3) can contain the SECOND pair of points P and Q
rewrite in standard form
y = 3 +x^2 + 2 x + 1
y = x^2 + 2 x + 4
I have no idea what the two original points P and Q were. They are on some other curve.
6 answers