Determine the vertex of the quadratic function y = 3x^2 - 5x + 2.
Answers:
To find the vertex of the quadratic function ( y = 3x^2 - 5x + 2 ), we can use the vertex formula for a parabola given by the standard form ( y = ax^2 + bx + c ).
The x-coordinate of the vertex can be found using the formula:
[ x = -\frac{b}{2a} ]
In this function, ( a = 3 ) and ( b = -5 ).
Now we can substitute the values of ( a ) and ( b ) into the formula:
[ x = -\frac{-5}{2 \times 3} = \frac{5}{6} ]
Next, we substitute ( x = \frac{5}{6} ) back into the original quadratic equation to find the y-coordinate of the vertex:
[ y = 3\left(\frac{5}{6}\right)^2 - 5\left(\frac{5}{6}\right) + 2 ]
Calculating ( \left(\frac{5}{6}\right)^2 ):
[ \left(\frac{5}{6}\right)^2 = \frac{25}{36} ]
Now substituting this back into the equation for ( y ):
[ y = 3 \cdot \frac{25}{36} - 5 \cdot \frac{5}{6} + 2 ]
Calculating ( 3 \cdot \frac{25}{36} ):
[ 3 \cdot \frac{25}{36} = \frac{75}{36} = \frac{25}{12} ]
Next, calculating ( 5 \cdot \frac{5}{6} ):
[ 5 \cdot \frac{5}{6} = \frac{25}{6} ]
Now we rewrite the equation for ( y ):
[ y = \frac{25}{12} - \frac{25}{6} + 2 ]
To combine these terms, we need a common denominator. The least common denominator of 12 and 6 is 12. Therefore, we rewrite ( \frac{25}{6} ):
[ \frac{25}{6} = \frac{50}{12} ]
Now substituting back into the equation:
[ y = \frac{25}{12} - \frac{50}{12} + 2 ]
This simplifies to:
[ y = \frac{25 - 50}{12} + 2 = -\frac{25}{12} + 2 ]
Converting 2 into twelfths:
[ 2 = \frac{24}{12} ]
Now we can combine these fractions:
[ y = -\frac{25}{12} + \frac{24}{12} = -\frac{1}{12} ]
Thus, the vertex of the quadratic function ( y = 3x^2 - 5x + 2 ) is:
[ \left(\frac{5}{6}, -\frac{1}{12}\right) ]