UpStudy Free Solution:

To find the slope of the curve at the given point \(P( 4, 36) \) and the equation of the tangent line at \(P\), we follow these steps:

1. Find the derivative of the function to get the slope of the tangent line.

The function is \(y = x^ 2 + 5x\). The derivative of \(y\) with respect to \(x\) is:

\[\frac { dy} { dx} = \frac { d} { dx} ( x^ 2 + 5x) = 2x + 5.\]

2. Evaluate the derivative at the given point to find the slope of the tangent line.

The given point is \(P( 4, 36) \). We substitute \(x = 4\) into the derivative:

\[\frac { dy} { dx} \bigg | _ { x= 4} = 2( 4) + 5 = 8 + 5 = 13.\]

So, the slope of the tangent line at \(P( 4, 36) \) is 13.

3. Use the point-slope form of the equation of a line to find the equation of the tangent line.

The point-slope form of a line is:

\[y - y_ 1 = m( x - x_ 1) ,\]

where \(m\) is the slope and \(( x_ 1, y_ 1) \) is the point on the line. Here, \(m = 13\) and \(( x_ 1, y_ 1) = ( 4, 36) \). Substituting these values in:

\[y - 36 = 13( x - 4) .\]

4. Simplify the equation to get it into slope-intercept form \(y = mx + b\).

\[y - 36 = 13( x - 4) \Rightarrow  y - 36 = 13x - 52 \Rightarrow  y = 13x - 52 + 36 \Rightarrow  y = 13x - 16.\]

Thus, the equation of the tangent line to the curve at the point \(P( 4, 36) \) is:

\[y = 13x - 16.\]

Therefore, the correct answer is:

D. slope is 13; \(y = 13x - 16\).

Supplemental Knowledge

In calculus, the slope of a curve at a given point is found using the derivative of the function. The derivative represents the rate of change of the function with respect to its variable. For a function \(y = f( x) \), the derivative \(f' ( x) \) gives us the slope of the tangent line to the curve at any point \(x\).

To find the equation of the tangent line at a specific point \(P( x_ 0, y_ 0) \), we use:

1. The slope of the tangent line, which is \(f' ( x_ 0) \).

2. The point-slope form of a line:

\[y - y_ 0 = m( x - x_ 0) \]

where \(m\) is the slope and \(( x_ 0, y_ 0) \) is the given point.

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