What is the minimum f(x)value for the following quadratic function? f(x)=2x^2+36x+170
+23254 A formula to find the x-value of the vertex of a quadratic function is: x = -b/(2a)
In f(x) = 2x² + 36x + 170 ---> a = 2, b = 36, and c = 170
x = -b/(2a) ---> x = -36/(2·2) = -36/4 = -9
f(-9) = 2(-9)² + 36(-9) + 170 = 8
Since the value of a > 0, this is a rising parabola, so the point (-9, 8) is a minimum.
The minimum f-value is 8.
+23254 A formula to find the x-value of the vertex of a quadratic function is: x = -b/(2a)
In f(x) = 2x² + 36x + 170 ---> a = 2, b = 36, and c = 170
x = -b/(2a) ---> x = -36/(2·2) = -36/4 = -9
f(-9) = 2(-9)² + 36(-9) + 170 = 8
Since the value of a > 0, this is a rising parabola, so the point (-9, 8) is a minimum.
The minimum f-value is 8.
