Parallel Tangents: Finding Points On Function Graphs

Hey guys! Let's dive into a super interesting problem in algebra: figuring out where the tangents to the graphs of two functions are parallel. This involves a bit of calculus (derivatives, specifically), but don't worry, we'll break it down step by step. We'll tackle this by finding the points where the derivatives of the functions are equal, since parallel lines have the same slope. So, let's get started!

Understanding Parallel Tangents

Before we jump into the specific examples, let's make sure we're all on the same page about what parallel tangents mean. Remember that the tangent to a curve at a point is a straight line that just touches the curve at that point. The slope of this tangent line is given by the derivative of the function at that point. So, if two curves have parallel tangents at certain points, it means their derivatives (slopes of the tangent lines) are equal at those points. This is the key concept we'll be using throughout this explanation.

Think of it like two cars driving on a road. If their paths are parallel at a certain moment, it means they're heading in the same direction with the same inclination relative to the road. In mathematical terms, the direction is represented by the slope, and the 'road' is the x-axis on our graph. This visual analogy often helps to understand the underlying concept.

Why is this important? Well, understanding parallel tangents helps in various fields, from optimizing designs in engineering to predicting trends in economics. It's a fundamental concept in calculus that bridges the gap between functions and their geometric representation. Plus, it's a fantastic exercise in applying the rules of differentiation and equation-solving, solidifying your understanding of calculus principles.

Now, to put this concept into practice, we'll need to calculate the derivatives of the given functions. Recall that the derivative of a function $f(x)$ gives us the slope of the tangent line at any point $x$. The power rule, which states that the derivative of $x^n$ is $nx^{n-1}$, is going to be our best friend here, especially for polynomial functions. For linear functions, the derivative is simply the coefficient of $x$. Once we have the derivatives, we'll set them equal to each other and solve for $x$. These $x$ values will give us the points where the tangents are parallel. Then, we'll substitute these $x$ values back into the original functions to find the corresponding $y$ values. This will give us the exact points (x, y) where the tangents are parallel. Remember, a tangent line is a line that touches a curve at a single point, and its slope at that point is given by the derivative of the curve's function. This means that if two curves have parallel tangents, their derivatives at the points of tangency must be equal.

Solving the Specific Cases

Alright, let's roll up our sleeves and solve the cases you've presented. We'll go through each one step-by-step, showing you exactly how to find those points where the tangents are parallel.

Case 1: $f(x) = 2x^2 - 3x + 4$, $g(x) = 12x - 8$

In this first case, we've got a quadratic function, $f(x)$, and a linear function, $g(x)$. The first thing we need to do is find their derivatives. Remember, the derivative gives us the slope of the tangent line at any point.

• Finding the derivatives:

  • For $f(x) = 2x^2 - 3x + 4$, the derivative $f'(x)$ is $4x - 3$ (using the power rule).
  • For $g(x) = 12x - 8$, the derivative $g'(x)$ is simply $12$ (the coefficient of $x$ in a linear function).

• Setting the derivatives equal: Now, we set $f'(x)$ equal to $g'(x)$ to find the $x$-values where the tangents are parallel: $4x - 3 = 12$.

• Solving for x: Solving this equation, we get $4x = 15$, so x = rac{15}{4}. This is the x-coordinate of the point where the tangents are parallel.

• Finding the y-coordinates: To find the corresponding $y$-coordinates, we plug x = rac{15}{4} back into the original functions:

  • f(rac{15}{4}) = 2(rac{15}{4})^2 - 3(rac{15}{4}) + 4 = rac{225}{8} - rac{45}{4} + 4 = rac{225 - 90 + 32}{8} = rac{167}{8}
  • g(rac{15}{4}) = 12(rac{15}{4}) - 8 = 45 - 8 = 37

• The Point: So, the point where the tangents are parallel for these two functions is (rac{15}{4}, rac{167}{8}) for $f(x)$ and (rac{15}{4}, 37) for $g(x)$. Notice that the y-coordinates are different, which is perfectly normal. The important thing is that the tangents at these x-values have the same slope.

Case 2: $f(x) = 18x + 19$, $g(x) = -15x + 18$

Next up, we have two linear functions. This one's actually a bit simpler! Let's follow the same steps:

• Finding the derivatives:

  • For $f(x) = 18x + 19$, the derivative $f'(x)$ is $18$.
  • For $g(x) = -15x + 18$, the derivative $g'(x)$ is $-15$.

• Setting the derivatives equal: We set $f'(x)$ equal to $g'(x)$: $18 = -15$.

• Analyzing the result: Wait a minute! $18$ is never equal to $-15$. This means there's no solution for $x$. So, what does this tell us? It means that the tangents to these two linear functions are never parallel. This makes sense because linear functions are straight lines, and these two lines have different slopes (18 and -15), so they'll never have parallel tangents.

Case 3: $f(x) = 2x + 13$, $g(x) = 4x - 19$

Okay, let's keep going with two more linear functions. This should help solidify the process.

• Finding the derivatives:

  • For $f(x) = 2x + 13$, the derivative $f'(x)$ is $2$.
  • For $g(x) = 4x - 19$, the derivative $g'(x)$ is $4$.

• Setting the derivatives equal: We set $f'(x)$ equal to $g'(x)$: $2 = 4$.

• Analyzing the result: Just like in the previous case, $2$ is never equal to $4$. So, again, there's no solution. The tangents to these lines are never parallel because they have different slopes.

Case 4: $f(x) = 2x^3$, $g(x) = 4x^2$

Now, let's tackle one with higher powers of $x$. This will give us a chance to use the power rule a bit more extensively.

• Finding the derivatives:

  • For $f(x) = 2x^3$, the derivative $f'(x)$ is $6x^2$.
  • For $g(x) = 4x^2$, the derivative $g'(x)$ is $8x$.

• Setting the derivatives equal: We set $f'(x)$ equal to $g'(x)$: $6x^2 = 8x$.

• Solving for x: To solve this, we rearrange the equation: $6x^2 - 8x = 0$. We can factor out a $2x$: $2x(3x - 4) = 0$. This gives us two possible solutions: $x = 0$ and $3x - 4 = 0$ which means x = rac{4}{3}.

• Finding the y-coordinates: Now, we plug these $x$-values back into the original functions:

  • For $x = 0$:
    • $f(0) = 2(0)^3 = 0$
    • $g(0) = 4(0)^2 = 0$ So, one point is $(0, 0)$.
  • For x = rac{4}{3}:
    • f(rac{4}{3}) = 2(rac{4}{3})^3 = 2(rac{64}{27}) = rac{128}{27}
    • g(rac{4}{3}) = 4(rac{4}{3})^2 = 4(rac{16}{9}) = rac{64}{9}

• The Points: So, the points where the tangents are parallel for these two functions are $(0, 0)$ and (rac{4}{3}, rac{128}{27}) for $f(x)$, and $(0, 0)$ and (rac{4}{3}, rac{64}{9}) for $g(x)$.

Case 5: $f(x) = ...$

Okay, so it looks like the fifth case is incomplete. If you provide the full function for $f(x)$, I’d be happy to walk you through the solution!

Key Takeaways and Next Steps

Wow, we've covered a lot! Finding points where tangents are parallel involves a pretty straightforward process:

1. Find the derivatives of both functions.
2. Set the derivatives equal to each other.
3. Solve for x. These are the x-coordinates where the tangents might be parallel.
4. Plug the x-values back into the original functions to find the corresponding y-coordinates.

The points you find are where the tangents have the same slope, meaning they're parallel. This concept is super useful in calculus and has applications in various fields.

To further your understanding, try graphing these functions and their tangent lines at the points you've found. This will give you a visual confirmation that the tangents are indeed parallel. You can use online graphing tools like Desmos or Geogebra for this. Also, try creating your own examples with different types of functions (trigonometric, exponential, etc.) to practice finding parallel tangents in different contexts.

And remember, calculus is all about practice! The more problems you solve, the more comfortable you'll become with the concepts. Don't hesitate to review the basic rules of differentiation and equation-solving if you need a refresher. Keep up the great work, guys, and you'll be a calculus pro in no time!