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Find the vertex of the function f (x) = x^2+6x-2

Sagot :

f(x)=x^2+6x-2, vertex: (-3, -11)
First, put -(6)/2(1) or -b/2a.
You then get -3 as the X value of the vertex.
Next, insert -3 back into the equation to get the Y value of the vertex.
(-3)^2+6(-3)-2 = -11
You then have -11 as the Y value of the vertex.


Panoyin
X - Vertex
[tex]-\frac{6}{2(1)} = -\frac{6}{2} = -3[/tex]

Y - Vertex
[tex]f(x) = x^{2} + 6x - 2 \\f(-3) = (-3)^{2} + 6(-3) - 2 \\f(-3) = 9 - 18 - 2 \\f(-3) = -9 - 2 \\f(-3) = -11[/tex]

Vertex
(x, y) = (3, -11)
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