Question
upstudy study bank question image url

7. Given that \( y=(2 x-3)(x+5)^{5} \), find the \( x \)-coordinates of the points where the tangent is perpendicular to the \( y \)-axis.

Ask by Romero Murray. in Singapore
Jan 27,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The tangent is perpendicular to the y-axis at \( x = -5 \) and \( x = \frac{5}{12} \).

Solution

To find the \( x \)-coordinates where the tangent to the curve \( y = (2x - 3)(x + 5)^5 \) is perpendicular to the \( y \)-axis, we need to determine where the slope of the tangent line is zero. A tangent line perpendicular to the \( y \)-axis is horizontal, meaning its slope is zero. ### Step 1: Differentiate \( y \) with respect to \( x \) Using the product rule for differentiation: \[ y = (2x - 3)(x + 5)^5 \] \[ \frac{dy}{dx} = (2x - 3) \cdot 5(x + 5)^4 + 2 \cdot (x + 5)^5 \] \[ \frac{dy}{dx} = 5(2x - 3)(x + 5)^4 + 2(x + 5)^5 \] ### Step 2: Factor the derivative Factor out the common term \( (x + 5)^4 \): \[ \frac{dy}{dx} = (x + 5)^4 [5(2x - 3) + 2(x + 5)] \] \[ \frac{dy}{dx} = (x + 5)^4 [10x - 15 + 2x + 10] \] \[ \frac{dy}{dx} = (x + 5)^4 (12x - 5) \] ### Step 3: Set the derivative equal to zero For the tangent to be horizontal: \[ (x + 5)^4 (12x - 5) = 0 \] This gives two solutions: 1. \( x + 5 = 0 \) \(\Rightarrow\) \( x = -5 \) 2. \( 12x - 5 = 0 \) \(\Rightarrow\) \( x = \frac{5}{12} \) ### Conclusion The \( x \)-coordinates where the tangent is perpendicular to the \( y \)-axis are: \[ x = -5 \quad \text{and} \quad x = \frac{5}{12} \] **Answer:** All real solutions are x = –5 and x = 5⁄12. The tangent is perpendicular to the y-axis at these x-values.

Reviewed and approved by the UpStudy tutoring team

error msg
Explain
Simplify this solution

Extra Insights

To find the \( x \)-coordinates of the points where the tangent is perpendicular to the \( y \)-axis, we first need to understand what this means in terms of the slope of the tangent line. A tangent that is perpendicular to the \( y \)-axis is horizontal, meaning its slope \( \frac{dy}{dx} \) must be zero. Start by differentiating \( y \) using the product and chain rules: 1. Let \( u = 2x - 3 \) and \( v = (x + 5)^5 \). 2. Use the product rule: \( \frac{dy}{dx} = u'v + uv' \). Calculate \( u' = 2 \), and \( v' = 5(x + 5)^4 \). Thus, we have: \[ \frac{dy}{dx} = (2)(x + 5)^5 + (2x - 3)(5(x + 5)^4) \] Now, set \( \frac{dy}{dx} = 0 \): \[ (2)(x + 5)^5 + (2x - 3)(5(x + 5)^4) = 0 \] Factor out \( (x + 5)^4 \): \[ (x + 5)^4 \left( 2(x + 5) + 5(2x - 3) \right) = 0 \] This gives two cases: 1. \( (x + 5)^4 = 0 \) → \( x + 5 = 0 \) → \( x = -5 \). 2. The term in parentheses \( 2(x + 5) + 5(2x - 3) = 0 \). Solving \( 2(x + 5) + 10x - 15 = 0 \): \[ 12x - 5 = 0 \implies x = \frac{5}{12} \] So the \( x \)-coordinates where the tangent is perpendicular to the \( y \)-axis are \( x = -5 \) and \( x = \frac{5}{12} \).

Related Questions

Multiple Choice Identify the choice that best completes the statement or answers the question. Find any points of discontinuity for the rational function. 1. \( y=\frac{(x-7)(x+2)(x-9)}{(x-5)(x-2)} \) a. \( x=-5, x=-2 \) b. \( x=5, x=2 \) c. \( x=-7, x=2, x=-9 \) d. \( x=7, x=-2, x=9 \) 2. \( y=\frac{(x+7)(x+4)(x+2)}{(x+5)(x-3)} \) a. \( x=-5, x=3 \) b. \( x=7, x=4, x=2 \) c. \( x=-7, x=-4, x=-2 \) d. \( x=5, x=-3 \) 3. \( y=\frac{x+4}{x^{2}+8 x+15} \) a. \( x=-5, x=-3 \) b. \( x=-4 \) c. \( x=-5, x=3 \) d. \( x=5, x=3 \) 4. \( y=\frac{x-3}{x^{2}+3 x-10} \) a. \( x=-5, x=2 \) b. \( x=5, x=-2 \) c. \( x=3 \) d. \( x \) \( =-5, x=-2 \) 6. What are the points of discontinuity? Are they all removable? \[ y=\frac{(x-4)}{x^{2}-13 x+36} \] a. \( x=-9, x=-4, x=8 \); yes b. \( x=1, x=8, x= \) -8; no c. \( x=9, x=4 \); no d. \( x=-9, x=-4 \); no 7. Describe the vertical asymptote(s) and hole(s) for the graph of \( y=\frac{(x-2)(x-5)}{(x-5)(x+2)} \). a. asymptote: \( x=2 \) and hole: \( x=-5 \) b. asymptotes: \( x=-2 \) and hole: \( x=-5 \) c. asymptote: \( x=-2 \) and hole: \( x=5 \) d. asymptote: \( x=-2 \) and hole: \( x=-2 \) a. \( x=-3, x=-8 \); no b. \( x=5, x=-7, x=1 \); no c. \( x=-5, x=7, x=-1 \); yes d. \( x=3, x=8 \); yes 8. Describe the vertical asymptote(s) and hole(s) for the graph of \( y=\frac{x+1}{x^{2}+6 x+5} \). a. asymptote: \( x=-1 \) and hole: \( x=-1 \) b. asymptote: \( x=-1 \) and hole: \( x=-1 \) c. asymptotes: \( x=-1,-1 \) and hole: \( x=-1 \) d. asymptote: -5 and hole: \( x=-1 \), 9. Find the horizontal asymptote of the graph of \( y=\frac{7 x^{6}+7 x+3}{9 x^{5}+7 x+3} \). a. \( y=0 \) b. \( y=\frac{7}{9} \) c. no horizontal asymptote d. \( y=\frac{6}{5} \)
Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy