Solved on Nov 08, 2023

Find the minimum value of the function y=3x212x+10y = 3x^2 - 12x + 10.

STEP 1

Assumptions1. The function is y=3x12x+10y=3x^{}-12x+10 . We are looking for the minimum value of the function

STEP 2

The given function is a quadratic function of the form y=ax2+bx+cy=ax^{2}+bx+c, where a=a=, b=12b=-12, and c=10c=10. The minimum or maximum value of a quadratic function occurs at the vertex of the parabola.

STEP 3

The x-coordinate of the vertex of a parabola given by y=ax2+bx+cy=ax^{2}+bx+c is given by b2a-\frac{b}{2a}.

STEP 4

Plug in the values for aa and bb to find the x-coordinate of the vertex.
xvertex=1223x_{vertex} = -\frac{-12}{2 \cdot3}

STEP 5

Calculate the x-coordinate of the vertex.
xvertex=1223=2x_{vertex} = -\frac{-12}{2 \cdot3} =2

STEP 6

Now that we have the x-coordinate of the vertex, we can find the y-coordinate by substituting xvertexx_{vertex} into the function.
yvertex=3(2)2122+10y_{vertex} =3 \cdot (2)^{2} -12 \cdot2 +10

STEP 7

Calculate the y-coordinate of the vertex.
yvertex=3(2)2122+10=2y_{vertex} =3 \cdot (2)^{2} -12 \cdot2 +10 = -2The minimum value of the function y=3x212x+10y=3x^{2}-12x+10 is 2-2.

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