Question

hey all help me plz
find the tangent line to f(x)=15-2x^2 at x=1

Answer 1

The point would be (1,13) solving for when x=1.
Slope would be -4 from taking derivative of f(x). You now have slope and a point. You have the info write the equation for the tangent line. Can you take it from here?

Answer 2

i attachments you > i want solve Question No. 4 if u can

Answer 3

The B Part, as A. was what was above.

Answer 4

oky the part A. solve it
f(x)=15-2x^2 at x=1
f(x)=15-2(1)^2
f(x)=15-2
f(x)= 13
The point would be (1,13) solving for when x=1.
Slope would be -4 from taking derivative of f(x).
thats true ? and how can Draw The Graph

and Part C and D ? can u solve it plz

Answer 5

B.\[-(x ^{3}-8)\over x ^{2}-4\] Please note, that the numerator is the difference of two perfect cubes, and the denominator is the difference of two perfect squares. These a special cases and can be factored as follows:\[-(x-2)(x ^{2}+2x+4)\over (x+2)(x-2)\]Now you can do some canceling and get:\[-(x ^{2}+2x+4 \over x+2\]The limit when x approaches 2 would be:\[12\over4\] or 3. B. is 3

Answer 6

Excuse me I forgot to consider the minus sign. -3 is the answer

Answer 7

oky thank you so so much

oky the part A. solve it
f(x)=15-2x^2 at x=1
f(x)=15-2(1)^2
f(x)=15-2
f(x)= 13
The point would be (1,13) solving for when x=1.
Slope would be -4 from taking derivative of f(x).
thats true ? and how can Draw The Graph

and Part C and D ? can u solve it plz

Answer 8

C. I don't understand the problem, just trying to figure it out
i. For a = -3 there is a break and f(a) becomes 4
ii. For a =-1, f(a) breaks and becomes a 3
iii. For a= 2, f(a) does not exist
iv. For a = 4, f(a)=5

Really am treading in unfamiliar waters here, so these are just SWAGslol
Hi robtobey, thigs ok in sunny CA

Answer 9

ummmm oky thx for u

but how can i draw the tangent line for part A (q.4)

and do u have solve the part D ( q.4)

am sorry

Answer 10

The line would be:

-4=(13-y)/(1-x)
-13y=4x-9
y=(-4/13)x+9/13