Find the vertex
H(x)= -3x^2+24x-46
What is the method you were taught for this procedure, since there are several ways to do this?
Example: 4x^2+8x-3
Y=4(x^2+8x+1)^2-3-4
Y=4(x+1)^2-7
Vertex:(-1,-7)
-3x^2+24x-46
=-3(x^2-8+16) +2
=-3(x-4)^2+2
x=4,y=2
H(x)= -3x^2+24x-46
= -3(x^2 - 8x + ....) - 46
= -3(x^2 - 8x + 16) - 46 + 48
= -3(x - 4)^2 + 2
vertex is (4, 2)
or
for f(x) = ax^2 + bx + c
the x of the vertex is -b/(2a)
for H(x)= -3x^2+24x-46
x of the vertex is -24/-6 = 4
H(4) = -48 + 96 - 46 = 2
vertex is (4,2)
To find the vertex of a quadratic function in the form h(x) = ax^2 + bx + c, you can use the formula x = -b/2a. The x-coordinate of the vertex is given by the formula x = -b/2a.
In the given quadratic function h(x) = -3x^2 + 24x - 46, we can identify that a = -3, b = 24, and c = -46.
To find the x-coordinate of the vertex, we substitute these values into the formula:
x = -b/2a
x = -24/(2*(-3))
x = -24/(-6)
x = 4
So, the x-coordinate of the vertex is x = 4.
To find the y-coordinate of the vertex, substitute the value of x into the given function:
h(x) = -3x^2 + 24x - 46
h(4) = -3(4)^2 + 24(4) - 46
h(4) = -3(16) + 96 - 46
h(4) = -48 + 96 - 46
h(4) = 2
Therefore, the y-coordinate of the vertex is y = 2.
Hence, the vertex of the given quadratic function H(x) = -3x^2 + 24x - 46 is (4, 2).