I'm supposed to make f(x)= 3x^2 - 18x +28 into vertex form
I got f(x) = 3(x - 3)^2 + 1
My question what would be the vertex?
Would it be (3,1) or (-3,1) I'm confused by the - in front of the 3
In the form
y = a(x-h)^2 + k
the vertex is (h,k), the value of a does not in any way affect with the
vertex
in yours, had the equation been y = 5(x-3)^2 + 1
the vertex would still be (3,1), the 5 has nothing to do with the vertex.
I was so more asking about the - inside the parenthesis. Like the (x-3) so would you not also have to take the - in front of the 3?
To find the vertex of a quadratic function in vertex form, we can use the formula x = -b/2a.
In the given quadratic function f(x) = 3x^2 - 18x + 28, we can see that a = 3 and b = -18.
Using the formula to find the x-coordinate of the vertex, we have:
x = -(-18) / (2 * 3) = 18 / 6 = 3
Therefore, the x-coordinate of the vertex is 3.
To find the y-coordinate of the vertex, we substitute the x-coordinate back into the function equation:
f(3) = 3(3)^2 - 18(3) + 28
= 9 - 54 + 28
= -17
Hence, the vertex of the quadratic function f(x) = 3x^2 - 18x + 28 in vertex form is (3, -17).
In the vertex form you obtained, f(x) = 3(x - 3)^2 + 1, the x-coordinate of the vertex is still 3. The -3 inside the parentheses represents a horizontal shift of the graph to the right by 3 units from the standard form. So, in this case, the vertex remains (3, -17) and is not affected by the -3 in front of the squared term. The y-coordinate of the vertex is given by the constant term outside the parentheses, which is 1. Hence, the vertex in this case is indeed (3, 1). The negative sign in front of the 3 does not affect the y-coordinate of the vertex.