Find the vertex of y = −3(x + 1)2 + 5
Find the vertex of y = −2t2 + 12t − 23
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
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.
To find the vertex of a quadratic equation in the form y = ax^2 + bx + c, you can use the formula x = -b/2a for the x-coordinate of the vertex.
For the first equation y = −3(x + 1)2 + 5, we can see that a = -3, b = -6, and c = 5. Plugging these values into the formula, we get x = -(-6)/(2*(-3)) = 6/(-6) = -1.
To find the y-coordinate of the vertex, substitute the x value (-1) into the equation to get y = -3(-1 + 1)^2 + 5 = -3(0)^2 + 5 = -3(0) + 5 = 5.
Therefore, the vertex of y = −3(x + 1)2 + 5 is (-1, 5).
For the second equation y = −2t^2 + 12t − 23, we can see that a = -2, b = 12, and c = -23. Using the formula x = -b/2a, we have x = -12/(2*(-2)) = 12/4 = 3.
Substituting the x value (3) into the equation to find the y-coordinate, we get y = -2(3)^2 + 12(3) - 23 = -2(9) + 36 - 23= -18 + 36 - 23 = -5.
Therefore, the vertex of y = −2t^2 + 12t − 23 is (3, -5).