Determining an Equation of a Tangent Line Prompt \#3; Consider the function \( f(x)=-3 x^{2}+6 x+4 \). Use whichever method you prefer from the previous two slides to determine the slope-intercept form of the equation of the tangent line to the graph of \( y=f(x) \) at \( x=2 \). Question \#3(a): What is the value of \( x_{0} \) ? Question \#3(b): What is the slope-intercept form of the equation of the tangent line? \( y= \)

To find the equation of the tangent line to the function \( f(x) = -3x^2 + 6x + 4 \) at \( x = 2 \), we need to follow these steps: ### Question \#3(a): What is the value of \( x_{0} \)? The value of \( x_{0} \) is the point at which we want to find the tangent line. In this case, \( x_{0} = 2 \). ### Question \#3(b): What is the slope-intercept form of the equation of the tangent line? 1. **Find \( f(2) \)**: \[ f(2) = -3(2)^2 + 6(2) + 4 \] \[ = -3(4) + 12 + 4 \] \[ = -12 + 12 + 4 = 4 \] So, the point on the graph at \( x = 2 \) is \( (2, 4) \). 2. **Find the derivative \( f'(x) \)** to determine the slope of the tangent line: \[ f'(x) = \frac{d}{dx}(-3x^2 + 6x + 4) = -6x + 6 \] 3. **Evaluate the derivative at \( x = 2 \)** to find the slope: \[ f'(2) = -6(2) + 6 = -12 + 6 = -6 \] Thus, the slope of the tangent line at \( x = 2 \) is \( -6 \). 4. **Use the point-slope form of the equation of a line**: The point-slope form is given by: \[ y - y_1 = m(x - x_1) \] where \( m \) is the slope and \( (x_1, y_1) \) is the point on the line. Here, \( m = -6 \) and \( (x_1, y_1) = (2, 4) \): \[ y - 4 = -6(x - 2) \] 5. **Simplify to slope-intercept form**: \[ y - 4 = -6x + 12 \] \[ y = -6x + 12 + 4 \] \[ y = -6x + 16 \] Thus, the slope-intercept form of the equation of the tangent line is: \[ \boxed{y = -6x + 16} \]