Finding the X-Coordinate of the Turning Point for ( f(x) -3x^2 2x 7 )

The question at hand is to find the x-coordinate of the turning point for the quadratic function ( f(x) -3x^2 2x 7 ). This is a fundamental concept in both algebra and calculus, and understanding how to find such points is crucial for tackling more complex mathematical problems.

Understanding the Basics

First, let's clarify the function:

Incorrect Approach:
-3x^2 2x 7 -3x^3 - 2/3x 7 -3x^2 - 2/3x 1/9 - 1/9 7 -3x - 1/3^2 1/3 7.
-3x - 1/3^2 22/3.
The turning point is 1/3 22/3.

Correct Approach:
y -3 [ x - (1/3)^2 - (22/9) ] -3 [ x - (1/3)^2 - (22/9) ]
-3x - (1/3)^2 y - (22/3)
x - (1/3)^2 (4 -1/12)y - (22/3)
The solution is 1/3 22/3.

Using Differential Calculus

For a quadratic equation of the form ( ax^2 bx c ), the x-coordinate of the turning point (vertex) can be found using the formula:

( x -frac{b}{2a} )

Given ( f(x) -3x^2 2x 7 ):

( a -3, , b 2 )

( x -frac{b}{2a} -frac{2}{2(-3)} frac{1}{3} )

Differentiation Method

The derivative of a function provides the slope at any point. At the turning point, the slope is zero. Let's differentiate ( f(x) ):

( f(x) -3x^2 2x 7 )

( frac{dy}{dx} -6x 2 )

Setting the derivative to zero to find the turning point:

( 0 -6x 2 )

( 6x 2 )

( x frac{2}{6} frac{1}{3} )

General Quadratic Function

For a general quadratic function, ( f(x) ax^2 bx c ), the x-coordinate of the turning point is given by:

( x_m frac{-b}{2a} )

For the function ( f(x) -3x^2 2x 7 ):

( a -3, , b 2 )

( x_m frac{-2}{2(-3)} frac{1}{3} )

Delta and Turning Point Coordinates

The turning point coordinates can also be derived using the discriminant ( Delta ) of the quadratic equation:

( Delta b^2 - 4ac )

( Delta 2^2 - 4(-3)(7) 4 84 88 )

The x-coordinate of the turning point is then given by:

( x frac{-b}{2a} frac{-2}{2(-3)} frac{1}{3} )

Thus, the turning point has coordinates ( left(frac{1}{3}, frac{22}{3}right) ).