What is the vertex given the equation: y=2/3 square root of x+4-2
Vertex:
our function is
y=23x+4−2y=\tfrac{2}{3}\sqrt{\smash[b]{x+4}}-2y=32x+4−2
Standard form of the square root function is
y=ab(x−h)+ky=a\sqrt{\smash[b]{b(x-h)}}+ky=ab(x−h)+k , where
b=±1b=±1b=±1 and
h,kh, kh,k - vertex
Convert our function
y=23x+4−2=23−(−x−4)−2y=\tfrac{2}{3}\sqrt{\smash[b]{x+4}}-2=\tfrac{2}{3}\sqrt{\smash[b]{-(-x-4)}}-2y=32x+4−2=32−(−x−4)−2
Therefore vertex is (−4,−2)(-4, -2)(−4,−2)
Also we can see vertex on the graph of the function.
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