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(a) Find the equation of the tangent to the curve \(y = x^ 3 + 2x^ 2 - 3x + 4\) at the point where the curve crosses the y-axis.

1. Find the point where the curve crosses the y-axis:

The curve crosses the y-axis where \(x = 0\).

Substituting \(x = 0\) into the equation of the curve:

\[y = 0^ 3 + 2( 0) ^ 2 - 3( 0) + 4 = 4\]

So, the point is \(( 0, 4) \).

2. Find the derivative of the curve to get the slope of the tangent:

The derivative of \(y = x^ 3 + 2x^ 2 - 3x + 4\) is:

\[\frac { dy} { dx} = 3x^ 2 + 4x - 3\]

3. Evaluate the derivative at \(x = 0\) to find the slope of the tangent:

\[\left . \frac { dy} { dx} \right | _ { x= 0} = 3( 0) ^ 2 + 4( 0) - 3 = - 3\]

So, the slope of the tangent at \(( 0, 4) \) is \(- 3\).

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

The point-slope form is \(y - y_ 1 = m( x - x_ 1) \), where \(m\) is the slope and \(( x_ 1, y_ 1) \) is the point.

Substituting \(m = - 3\) and \(( x_ 1, y_ 1) = ( 0, 4) \):

\[y - 4 = - 3( x - 0) \]

Simplifying:

\[y = - 3x + 4\]

So, the equation of the tangent is \(y = - 3x + 4\).

(b) Find the coordinates of the point where the tangent meets the curve again.

1. Set the equation of the curve equal to the equation of the tangent:

\[x^ 3 + 2x^ 2 - 3x + 4 = - 3x + 4\]

2. Simplify the equation:

\[x^ 3 + 2x^ 2 - 3x + 4 + 3x - 4 = 0\]

\[x^ 3 + 2x^ 2 = 0\]

3. Factor the equation:

\[x^ 2( x + 2) = 0\]

So, \(x^ 2 = 0\) or \(x + 2 = 0\).

This gives us \(x = 0\) or \(x = - 2\).

4. Find the corresponding \(y\)-coordinates:

For \(x = 0\), we already know \(y = 4\).

For \(x = - 2\):

\[y = ( - 2) ^ 3 + 2( - 2) ^ 2 - 3( - 2) + 4\]

\[y = - 8 + 8 + 6 + 4\]

\[y = 10\]

So, the coordinates of the point where the tangent meets the curve again are \(( - 2, 10) \).

Summary

(a) The equation of the tangent to the curve at the point where it crosses the y-axis is \(y = - 3x + 4\).

(b) The coordinates of the point where the tangent meets the curve again are \(( - 2, 10) \).

Supplemental Knowledge

Understanding how to find the equation of a tangent line to a curve at a given point is a fundamental concept in calculus. This involves several steps:

1. Finding the Point of Tangency:

- Identify the point on the curve where you need to find the tangent. This often involves substituting specific values into the function.

2. Calculating the Derivative:

- The derivative of a function gives us the slope of the tangent line at any point on the curve. For a function \(y = f( x) \), its derivative \(\frac { dy} { dx} \) or \(f' ( x) \) represents this slope.

3. Evaluating the Derivative at the Point of Tangency:

- Substitute the x-coordinate of your point into the derivative to find the slope at that specific point.

4. Using Point-Slope Form for Line Equation:

- With the slope and coordinates of your point, use the point-slope form \(y - y_ 1 = m( x - x_ 1) \), where \(m\) is your slope and \(( x_ 1, y_ 1) \) is your point, to write out your tangent line equation.

5. Finding Intersection Points:

- To find where this tangent intersects with its original curve again, set their equations equal and solve for x-coordinates, then substitute back to find corresponding y-coordinates.

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