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Solved on Nov 08, 2023

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

STEP 1

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

STEP 2

The given function is a quadratic function of the form $y=ax^{2}+bx+c$ , where $a=$ , $b=-12$ , and $c=10$ . 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 $y=ax^{2}+bx+c$ is given by $-\frac{b}{2a}$ .

STEP 4

Plug in the values for $a$ and $b$ to find the x-coordinate of the vertex.
$x_{vertex} = -\frac{-12}{2 \cdot3}$

STEP 5

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

STEP 6

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

STEP 7

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

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