Here's my question :) Thanks!
Let f(x) = a (7-x^2) for a ≠ 0
a0 Find the equations of the lines tangent to these curves at x=-1
b) find, in terms of a, the y-intercepts of the tangetn lines at x=-1
c) find the x-intercepts of the tangent lines at x=-1
d) Find, in terms of a, the area enclosed by the graph of f(x), the tangent line at x=-1, and the y-axis.
I solved part a and got y=2ax + 8a which i know is right
i solved part by and i got 8a
Part c i'm so sure about is it +/- the square root of 7? Do i set y to zero in the original or derivative??
Part d i have no clue about so please help!
I'm looking for help on parts c and d! THANK YOU SO MUCH :)
Update:@ Old Teacher- thank you so much!!! I tried part c by myself before i saw this and i got x=-4 as well so thank you! For part D, i'm still a little confused on these steps:
1)= ax^3/3 + ax^2+ ax| (-1,0)
2) =0-( -a/3 +a-a)
3 )=a/3
how exactly did you do that and make it ax^3/3? I understand everything up to those last steps. thanks again! :)
Answers & Comments
Verified answer
f(x)= 7a-ax^2
f '(x)= -2ax
A)
At x=-1, y= 6a
m= 2a
Y= 2ax + 8a (tangent line is correct)
B) 2a(0) +8a= 8a (correct for y intercept)
C) x-intercept of the tangent line is where y is zero.
0= 2ax +8a
X=-4
D) integrate (tangent line- function) on [-1,0]
Edit: the line is above the function if a is positive:
INT[ 2ax+ 8a)- (-ax^2+7a) ] dx on [-1,0]
= INT[ ax^2+2ax +a]dx
Treat a as a constant : (integrate as if a was 3)
= ax^3/3 + ax^2+ ax| (-1,0)
Now plug in 0 and -1 for x:
=0-( -a/3 +a-a)
=a/3
Hoping this helps!