Help 18 cubic function whose graph has horizontal tangents at (-2,6) and (2₁0). {Y= ax ³ + bx²² textd 0-(-2,6) 1+ (20) Y= ax²+bx³²+cxtd (Y= ax²³² +bx²³² textd 6= a(-2) +b(-2)² + c(-2) +d (0=a₂ (2) ²³+ b(2)+c(2) + d 6=-8a + 4b-2 ctd ⒸY'(x) = ax²+bx² +cx +d (/(-2)=3arc²³² + 2 bx+c 10=8a + 4b + 2 c t d 3a (-2)² + 2b (-2) +c =a (3x²)+b(2x) + c(1) = 3ax²³² +2bx+c Ⓒf(2)=3ax²²³ +2bx+c 3a (2)² + 2b (2) tr 12a + 4b tc 4b+d+ 4b+d= 6 8b+2d=6 8(0) + 2d=6 2d=6 용 d=34 98a+ 4b +2c td = 0 d= 216046-26-OSY 79, (TO) 8a+4b-2+d+86 +46 +2c+d=6) 2b =0) (²x + Nook 12a + 2b +c (ƒ (2) - f (-2) E- 12a + 4b+c-412a + 2b + c) = 0 1₁2a + 4bte - sza-2bx=0 Ba +4(0) + 2 (-12a) +3=0 8a-24a +3=0 1-16 a=-3 -3 a=-16 3 a = 16 +b=0b=0 12a + 2 b + c =0 12a + 2(0)+c=0 12a+c=0 =-120 (r = -12a² = ² + 1) = (x) \ | c= -12 (²76) 36 |C=76 C==4 3 Y=ar the textd 3 3 • + (0)²³²+ ( a hc +3 1670 339 16x² - 4x +3 O E
Help 18 cubic function whose graph has horizontal tangents at (-2,6) and (2₁0). {Y= ax ³ + bx²² textd 0-(-2,6) 1+ (20) Y= ax²+bx³²+cxtd (Y= ax²³² +bx²³² textd 6= a(-2) +b(-2)² + c(-2) +d (0=a₂ (2) ²³+ b(2)+c(2) + d 6=-8a + 4b-2 ctd ⒸY'(x) = ax²+bx² +cx +d (/(-2)=3arc²³² + 2 bx+c 10=8a + 4b + 2 c t d 3a (-2)² + 2b (-2) +c =a (3x²)+b(2x) + c(1) = 3ax²³² +2bx+c Ⓒf(2)=3ax²²³ +2bx+c 3a (2)² + 2b (2) tr 12a + 4b tc 4b+d+ 4b+d= 6 8b+2d=6 8(0) + 2d=6 2d=6 용 d=34 98a+ 4b +2c td = 0 d= 216046-26-OSY 79, (TO) 8a+4b-2+d+86 +46 +2c+d=6) 2b =0) (²x + Nook 12a + 2b +c (ƒ (2) - f (-2) E- 12a + 4b+c-412a + 2b + c) = 0 1₁2a + 4bte - sza-2bx=0 Ba +4(0) + 2 (-12a) +3=0 8a-24a +3=0 1-16 a=-3 -3 a=-16 3 a = 16 +b=0b=0 12a + 2 b + c =0 12a + 2(0)+c=0 12a+c=0 =-120 (r = -12a² = ² + 1) = (x) \ | c= -12 (²76) 36 |C=76 C==4 3 Y=ar the textd 3 3 • + (0)²³²+ ( a hc +3 1670 339 16x² - 4x +3 O E
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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### Finding a Cubic Function with Given Properties
We are given a **cubic function with horizontal tangents at the points (2,6) and (-2,6)**. Our goal is to determine this cubic function. The general form of a cubic function is:
\[ y = ax^3 + bx^2 + cx + d \]
1. Plugging in the points (2,6) and (-2,6), we get the following equations:
\[
y = ax^3 + bx^2 + cx + d
\]
- For \( x = 2, y = 6 \):
\[
6 = a(2)^3 + b(2)^2 + c(2) + d \implies 6 = 8a + 4b + 2c + d
\]
Rearrange to obtain Equation (1):
\[
8a + 4b + 2c + d = 6
\]
- For \( x = -2, y = 6 \):
\[
y = a(-2)^3 + b(-2)^2 + c(-2) + d \implies 6 = -8a + 4b - 2c + d
\]
Rearrange to obtain Equation (2):
\[
-8a + 4b - 2c + d = 6
\]
2. Since we have horizontal tangents at these points, the derivative \( y' \) at these points must equal 0:
\[
y' = 3ax^2 + 2bx + c
\]
- For \( x = 2, y' = 0 \):
\[
0 = 3a(2)^2 + 2b(2) + c \implies 0 = 12a + 4b + c
\]
Rearrange to obtain Equation (3):
\[
12a + 4b + c = 0
\]
- For \( x = -2, y' = 0 \):
\[
y' = 3a(-2)^2 + 2b(-2) + c \implies](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F42c8eea7-5a86-4f7e-acf0-00a343985dba%2Fc40419aa-4e28-4821-a5bc-1849987958d2%2Fxsewgw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:---
### Finding a Cubic Function with Given Properties
We are given a **cubic function with horizontal tangents at the points (2,6) and (-2,6)**. Our goal is to determine this cubic function. The general form of a cubic function is:
\[ y = ax^3 + bx^2 + cx + d \]
1. Plugging in the points (2,6) and (-2,6), we get the following equations:
\[
y = ax^3 + bx^2 + cx + d
\]
- For \( x = 2, y = 6 \):
\[
6 = a(2)^3 + b(2)^2 + c(2) + d \implies 6 = 8a + 4b + 2c + d
\]
Rearrange to obtain Equation (1):
\[
8a + 4b + 2c + d = 6
\]
- For \( x = -2, y = 6 \):
\[
y = a(-2)^3 + b(-2)^2 + c(-2) + d \implies 6 = -8a + 4b - 2c + d
\]
Rearrange to obtain Equation (2):
\[
-8a + 4b - 2c + d = 6
\]
2. Since we have horizontal tangents at these points, the derivative \( y' \) at these points must equal 0:
\[
y' = 3ax^2 + 2bx + c
\]
- For \( x = 2, y' = 0 \):
\[
0 = 3a(2)^2 + 2b(2) + c \implies 0 = 12a + 4b + c
\]
Rearrange to obtain Equation (3):
\[
12a + 4b + c = 0
\]
- For \( x = -2, y' = 0 \):
\[
y' = 3a(-2)^2 + 2b(-2) + c \implies
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