Write the equation of the tangent line to the curve at the indicated poin \[ y=3 x^{2}-5 x+7 \text { at } x=-3 \]

To find the equation of the tangent line to the curve \( y = 3x^2 - 5x + 7 \) at the point where \( x = -3 \), we will follow these steps: 1. **Find the y-coordinate of the point on the curve at \( x = -3 \)**. 2. **Calculate the derivative of the function to find the slope of the tangent line**. 3. **Use the point-slope form of the equation of a line to write the equation of the tangent line**. ### Step 1: Find the y-coordinate We substitute \( x = -3 \) into the equation of the curve: \[ y = 3(-3)^2 - 5(-3) + 7 \] Calculating this gives: \[ y = 3(9) + 15 + 7 = 27 + 15 + 7 = 49 \] So, the point on the curve is \( (-3, 49) \). ### Step 2: Calculate the derivative Next, we find the derivative of the function \( y = 3x^2 - 5x + 7 \) to determine the slope of the tangent line. The derivative \( y' \) is given by: \[ y' = \frac{d}{dx}(3x^2 - 5x + 7) \] Calculating the derivative: \[ y' = 6x - 5 \] Now, we evaluate the derivative at \( x = -3 \): \[ y'(-3) = 6(-3) - 5 = -18 - 5 = -23 \] So, the slope of the tangent line at the point \( (-3, 49) \) is \( -23 \). ### Step 3: Write the equation of the tangent line Using the point-slope form of the equation of a line, which 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, we substitute \( m = -23 \), \( x_1 = -3 \), and \( y_1 = 49 \): \[ y - 49 = -23(x + 3) \] Now, we can simplify this equation: \[ y - 49 = -23x - 69 \] \[ y = -23x - 69 + 49 \] \[ y = -23x - 20 \] Thus, the equation of the tangent line to the curve at the point \( x = -3 \) is: \[ \boxed{y = -23x - 20} \]