Question

Tangent lines:

Answer 1

http://prntscr.com/2vmbue

Answer 2

@wio

Answer 3

\[\Large\bf\sf y=ax^2\]Mmmm ok let's get a slope for our tangent line at x=2.

Answer 4

So take the derivative of y.

Answer 5

Then evaluate the derivative at x=2.

Answer 6

What's that give you? :o

Answer 7

that's
\[\LARGE y'=4a

Answer 8

*\[\LARGE y'=4a\]

Answer 9

\[\Large\bf\sf y'(2)\quad=\quad 4a\]Ok good.
Let's compare that result with our line that's been given.
(We'll put in slope-intercept form).\[\Large\bf\sf 2x+y=b \qquad\to\qquad y=-2x+b\]

Answer 10

The slope of this line is -2, yes?

Answer 11

Yup

Answer 12

So the line will be tangent to our curve when \(\Large\bf\sf -2\quad=\quad 4a\)

Answer 13

So a=-2

Answer 14

Ok good.
Now that we have our a value, we can plug x=2 into our original function to get a coordinate pair.

We can then use that coordinate pair to solve for b!

Answer 15

So:
(2, -8)

Answer 16

Ok cool, use that pair to find your b.\[\Large\bf\sf (2,\;-8)\qquad\to\qquad y=-2x+b\]

Answer 17

I just checked and a isn't -2 ._.

Answer 18

Oh oh we made a small boo boo earlier.

-2 = 4a --> a=-1/2.

Answer 19

The littlest things xD

Answer 20

ya :c

Answer 21

That means we need to recalculate our coordinate pair! :O

Answer 22

So.. \[\LARGE (-\frac{1}{2},~-\dfrac{1}{2})\]

Answer 23

I think it's -2 for the y coordinate.

Answer 24

\[\Large\bf\sf y(x)=ax^2 \qquad\to\qquad y(2)=-\frac{1}{2}(2)^2\quad=\quad -2\]

Answer 25

Oh, right..

Answer 26

Figure out that b value yet? c:

Answer 27

I thought I got it but now Idk ._.