UpStudy Free Solution:
Let's solve the problem step by step.
(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.
Mastering calculus concepts can greatly enhance your problem-solving capabilities! UpStudy can offer more in-depth explanations or assistance for challenging calculus problems with precise solutions and detailed step-by-step explanations across many subjects at all levels - everything from differentiation or exam prep. Our professional tutors are online 24/7 offering personalized assistance tailored just to you - elevating academic journey with UpStudy where every question finds an answer!