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Vertex form: \(y=a(x-h)^2+k\), where (h, k) is the vertex. Looking at the info they gave you, can you tell what either coordinate for the vertex is?
(-2, ? ) im sorry this is a new concept for me
no problem. The axis of symmetry gives the x-coordinate of the vertex. The maximum (or minimum when you get a problem like that) height gives the y-coordinate. So the vertex is (2, 2)
Put that into the equation to get \(y = a(x -2)^2+2\)
make sense so far?
because it is maximum, wont that make it a negative cooridinate or does that not apply here?
no. The negative doesn't come into play here. Just take the number as they give it to you. The negative comes in as a check. Since it's a maximum \(a\) is supposed to be negative. But you have to plug in the point (3, 0) to find it.
okay, i dont know how to plug in the points in runs through, im not really sure about that part
and yes, i was looking at my chart wrong, i see now that the a is whats supposed to be negative
You have \(y=a(x-2)^2+2\) You have to plug in 3 for x and 0 for y, then solve for a. \[0=a(3-2)^2+2\]
a is our maximum value which is 2, but its negative? so -2?
a is not the maximum value. The y-coordinate of the vertex is the maximum value.
ohhh ok, my bad
so the final answer is \[y=2(x-2)^2-2\]
so how does a=2 ?
no, it's not 2 \(0=a(3−2)^2+2\) \(0=a + 2\) \(-2=a\)
oh ok, i see what you did there. do you think you could help me with one more? (its similar, but worded differently, i will definitely contribute more this time, because i think i understand slightly better now...)