Question

What is the equation of the tangent line passing through the point (1, 3) of the graph of the function f(x) = x^2 + x + 1?

Answer 1

Are you using
\[\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}=f'(x)\]

or you using short cut methods ?

Answer 2

Both really

Answer 3

Use either the definition or the short cut methods to differentiate x^2+x+1

Answer 4

This will give you the slope formula...
This means this will give you the slope of the tangent line at any x

Answer 5

a line can with point (a,b) and slope m can be written as
y-b=m(x-a)

so we already know (a,b) which is (1,3)

we just need to find m which is f' evaluated at (1,3) or just x=1 since our function for f' will be in terms of x

Answer 6

m = (y-3)/(x-1)

Answer 7

and remember y is given as x^2+x+1
and you also forgetting to insert the limit as x approaches 1 there

Answer 8

Blah, my math is terrible and I'm not very good at limits, can we work through that?

Answer 9

do you know how to simplify x^2+x+1-3?

Answer 10

x^2+x-2 ?

Answer 11

yes do you know how to factor x^2+x-2?

Answer 12

(x+2)(x-1)

Answer 13

and what do you have on bottom there ?

Answer 14

you had (y-3)/(x-1)
as x approaches 1

Answer 15

you factored y-3 as (x+2)(x-1)

Answer 16

so you have
(x+2)(x-1)/(x-1)

as x approaches 1

Answer 17

since you know x-1 isn't 0
you can say (x-1)/(x-1)=?

Answer 18

5/5=?
6/6=?
-1/-1=?
-8/-8=?

...
(x-1)/(x-1)=? assuming x-1 isn't ever 0

Answer 19

something/(the same something)=1
when the something isn't zero

Answer 20

Sorry my tutor came over, it would be 1

Answer 21

yes so you are left with (x+2) as x approaches 1

Answer 22

\[y=x^2+x+1 \\ m=\lim_{x \rightarrow 1}\frac{y-3}{x-1}=\lim_{x \rightarrow 1} \frac{(x^2+x+1)-3}{x-1} \ \\ =\lim_{x \rightarrow 1}\frac{x^2+x-2}{x-1}=\lim_{x \rightarrow 1} \frac{(x+2)(x-1)}{x-1}=\lim_{x \rightarrow 1} (x+2)\]

Answer 23

do you know how to finish evaluating the slope there?