What is the minimum value of y if y=3x^2+8x+9+x+11?
y = 3x2 + 8x + 9 + x + 11
y = 3x2 + 9x + 20
take 1st derivative
and set equal to 0 6x + 9 = 0
6x = – 9
x = – 9/6 = – 1.5 <<— this is the x-coordinate of the vertex
plug –1.5 back in
original equation y = 3x2 + 9x + 20
y = 3(–1.5)2 + 9(–1.5) + 20
y = 6.75 – 13.5 + 20
y = 13.25 <<— this is the y-coordinate of the vertex
The minimum value of y is its value at the vertex = 13.25
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