Find the equation of tangent to curve `y=3x^(2) +4x +5` at `(0,5)`.

Answer

To find the equation of the tangent to the curve \( y = 3x^2 + 4x + 5 \) at the point \( (0, 5) \), we will follow these steps:

### Step 1: Write down the equation of the curve.
The equation of the curve is given as:
\[ y = 3x^2 + 4x + 5 \]

### Step 2: Find the derivative of the curve.
To find the slope of the tangent line at a given point, we need to compute the derivative \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{d}{dx}(3x^2 + 4x + 5) \]
Using the power rule:
\[ \frac{dy}{dx} = 6x + 4 \]

### Step 3: Evaluate the derivative at the point \( (0, 5) \).
Now, we will substitute \( x = 0 \) into the derivative to find the slope of the tangent line at that point:
\[ \frac{dy}{dx} \bigg|_{x=0} = 6(0) + 4 = 4 \]
Thus, the slope \( m \) of the tangent line at the point \( (0, 5) \) is \( 4 \).

### Step 4: Use the point-slope form to find the equation of the tangent.
The point-slope form of the equation of a line is given by:
\[ y - y_1 = m(x - x_1) \]
Here, \( (x_1, y_1) = (0, 5) \) and \( m = 4 \):
\[ y - 5 = 4(x - 0) \]
This simplifies to:
\[ y - 5 = 4x \]
So, adding \( 5 \) to both sides, we get:
\[ y = 4x + 5 \]

### Conclusion
The equation of the tangent to the curve at the point \( (0, 5) \) is:
\[ \boxed{y = 4x + 5} \]