Posted by Hokoslavia on Sunday, March 9, 2008 at 11:36am.
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.
Related Questions
math help,algebra - Okay this is what i have to do but i think i am doing ...
Algebra - Identify the vertex and the axis of symmetry for the graph of y=5(x-2...
Algebra - The formula is y=ax^2+bx+c I also gave the wrong problems, they are...
Algebra 2 - 3. Identify the vertex and the axis of symmetry of the graph of the ...
algebra (checking answer) - f(g)=(x-4)^2+6 Sketch a graph of f(x): 4 on y axis ...
math - A parabola has the equation y = 4(x-3)^2-7 Choose 2 true statements: A) ...
ALGEBRA 1 - Please check for me.Without drawing the graph of the given equation ...
ALGEBRA 1 - Please check for me.Without drawing the graph of the given equation ...
AlgebraB-2 - 1. What are the coordinates of the vertex of the graph? Is it a ...
algebra - I would like to have some help on this problem please. I'm so post...
For Further Reading