Find the vertex of y = −3(x + 1)^2 + 5
The vertex is where the function is highest in this case. The smallest number that gets subtracted from 5 is zero, and that is when the funtion is at its highest value. That happens when x = -1. The y value at that point is 5. Therefore the vertex is at (-1,5)
In the second example, rewrite it by adding AND subtracting 18 from the right side, so the "t" terms form a perfect square.
y = -2(t^2 -6t +9) -23 +18
y = -2(t-3)^2 -5
The vertex is at t = 3 and is a maximum. The y value there is -5
Find the vertex of y = −3(x + 1)2 + 5
Find the vertex of y = −2t2 + 12t − 23
3 answers
How did you get 6,9 and 18?
−2t^2 + 12t = 2(-t^2 +6t)
That is where the 6 came from
I got the 9 and the 18 in the process of completing the square
I suggest you review the subject of completing the square in quadratic equations.
That is where the 6 came from
I got the 9 and the 18 in the process of completing the square
I suggest you review the subject of completing the square in quadratic equations.
1 answer