Question

Hey so I thought I knew how to do this but I got the wrong answer so idk. Can someone give me a few pointers?

my answer was
L(x)= 7+ 3x^2-2(x-2)



**postin pic

Answer 1

1st take the derivative and then plug in x=2 into the result. This will give the slope of the line at that point of L(2). Next, using y=mx+b, where m = L'(2), which you just computed. Plug in x=2 into L(x) to find L(2). Then using this information, plug in y=L(2) , m=L'(2) and x=2 into y=mx+b to determine the value for b. Finally, you will have m and b, and the equation tangent to L(x).

Does this make sense?

Answer 2

oh yes it does thanks! I just realized I was finding the tangent line approximation... lol thank you!

Answer 3

your welcome

Answer 4

*you're

Answer 5

hey sorry but I got everything but b. so far my F'(2)= 10 (m), f(2)=7
so i got 7=10(2)+b. Meaning b is 7/20 but it's not right. Am i doing something wrong?

Answer 6

let me check your work
it will take a second

Answer 7

\(f(2)=7\) right?

Answer 8

\(f'(2)=10\) so you have \(m=10\) and the point \((2,7)\)
use the point slope formula

Answer 9

oooh lol
\(10=7\times 2+b=\iff 10=14+b\iff b=-4\)
not sure where the \(\frac{7}{20}\) came from

Answer 10

ohh lol i see! Wow i need wake up. I made 7 my y and 10 my slope so it looked like this:
7 = 10 x 2 + b

Answer 11

Thank you!

Answer 12

answer would be \(y=10x-4\)

Answer 13

yw

Answer 14

ok so i just checked my answer key and it's actually L(x)=10x-13 any idea how they got that answer?

Answer 15

The tangent passes through the point P(a,f(a)) with slope f'(a), so its point-slope equation is
$$
y-f(a)=f'(a)(x-a)
$$or
$$
y=f(a) + f'(a)(x-a)
$$
For us, \(f(2)=7\) and \(f'(2)=10\).
So, the tangent of the line is the graph of the function
$$L(x)=f(2)+f'(2)(x-2)\\
\qquad=7+10(x-2)\\
\qquad=7+10x-20\\
\qquad =10x-13$$