SOLUTION: if x is a real number find the minimum value of 3^x^2-4x+8 (x to power of 2, x-4 not 2-4) thanks for help

Question 1107754: if x is a real number find the minimum value of 3^x^2-4x+8 (x to power of 2, x-4 not 2-4)
thanks for help
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
y = 3x^2 - 4x + 8.

the min/max point on this quadratic equation is when x = -b/2a

the quadratic is in standard form of ax^2 + bx + c = 0 when y = 0.

therefore a = 3, b = -4, c = 8.

x = -b/2a becomes x = 4/6 = 2/3.

when x = 2/3, 3x^2 - 4x + 8 = 3 * 4/9 - 4 * 2/3 + 8 which is equal to 12/9 - 4 * 6/9 + 72/9 which is equal to 12/9 - 24/9 + 72/9 which is equal to 60/9.

since the coefficient of the x^2 term is positive, this equation will have a minimum value at (2/3, 60/9).

your minimum value for the equation is when x = 2/3 and y = 60/9.

in decimal form, that would be x = 0.667 and y = 6.667.

i graphed the equation of y = 3x^2 - 4x + 8 and the graph shows that the minimum point is at (.667,6.667), agreeing with the algebraic solution.

here's the graph.

$$$

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