Find the slope of the tangent to the curve `y=3x^(2)-5x+2` at `x=3.`

Answer

To find the slope of the tangent to the curve \( y = 3x^2 - 5x + 2 \) at \( x = 3 \), we will follow these steps:

### Step 1: Differentiate the function
To find the slope of the tangent line at a specific point, we first need to find the derivative of the function \( y \) with respect to \( x \).

The given function is:
\[ y = 3x^2 - 5x + 2 \]

Now, we differentiate \( y \):
\[ \frac{dy}{dx} = \frac{d}{dx}(3x^2) - \frac{d}{dx}(5x) + \frac{d}{dx}(2) \]

Using the power rule:
\[ \frac{dy}{dx} = 6x - 5 \]

### Step 2: Evaluate the derivative at \( x = 3 \)
Next, we need to find the slope of the tangent line at \( x = 3 \) by substituting \( x = 3 \) into the derivative we found.

\[ \frac{dy}{dx} \bigg|_{x=3} = 6(3) - 5 \]

Calculating this gives:
\[ \frac{dy}{dx} \bigg|_{x=3} = 18 - 5 = 13 \]

### Conclusion
Thus, the slope of the tangent to the curve at \( x = 3 \) is:
\[ \boxed{13} \]

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