[Algebra II] Parabolas?
I can't seem to find another point to use in this.. If anyone can give me one, with explanations on how to get it, that'd be great.
For each function, the vertex of the function's graph is given. Find a and b.
y = ax^2 + bx - 27; vertex is (2, -3).
Thanks in advance.
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Well if you are just looking for one other point...
you can always resort to the y-intercept (0,-27). That's a garanteed point on the graph.
Just in case you don't know... the y-intercept is the constant at the end of the equation think of y=mx+b the constant b is always the y-intercept.
Hope that helps!
I can't for the life of me remember how to do this, but I do know that if your line is crossing the y axis at -27 and the vertex is (2,-3) then a is going to be negative to show that the parabola is upside down. If I remember anything else I will be sure to let you know!