Question

Find all locations (x,y) on the graph of f(x)= x^2 +4x+3 where slope =10

Answer 1

@girondasearch @girondasearch @zepdrix

Answer 2

Using calculus techniques?

Answer 3

yes!

Answer 4

I'm in precalc

Answer 5

The derivative gives us slope at a particular point,
so we're looking for x values such that\[\large\rm f'(x)=10\]Do you understand how to find f'(x)?

Answer 6

calculate the derivative and set it equal to 10 :)

Answer 7

the slope is given which is 10

Answer 8

so they're asking the points where the slope i.e. derivative is 10

Answer 9

so @Ammarah do what @zepdrix said, but first find the derivative for us

Answer 10

do I use f(x +h)-f(x)/h?

Answer 11

there is a much easier way

Answer 12

can someone help me solve and graph this please

Answer 13

do i plug in points?

Answer 14

to take the derivative of a polynomial here's your formula you need to use

\[\frac{ d }{ dx }x^{n} = nx^{n-1}\]

Answer 15

what is n

Answer 16

so n is the exponent

Answer 17

\[x^{2}\] say if I had that can you tell me what n would be?

Answer 18

2

Answer 19

so using the formula above what would be the derivative of

\[x^{2}\]

Answer 20

idk

Answer 21

is ther another way? I never learned this formula

Answer 22

Just try, \[\frac{ d }{ dx }x^{n} = nx^{n-1}\]

Answer 23

like can i use point slope form or plug in points for x?

Answer 24

you have to follow me, because this is the fastest way

Answer 25

we take the exponent and we subtract 1 right?

so in x^{2} the exponent is 2. 2-1 = 1

then 2 goes in front of the x

\[\frac{ d }{ dx } x^{2} = 2x \]

Answer 26

@Ammarah can you see what the pattern is?

Answer 27

yes

Answer 28

then what do I do?

Answer 29

Oh you're using the limit definition of derivative?

Answer 30

\[\large\rm \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\]yes?

Answer 31

so I would just plug that in? how do I find all the locations?

Answer 32

Yes, that will give you the function which determines the locations.
Make sure you understand how to plug in the pieces,
it can be a little tricky.
\[\large\rm f(\color{orangered}{x})=(\color{orangered}{x})^2+4(\color{orangered}{x})+3\]So the other piece in the numerator is going to be,\[\large\rm f(\color{orangered}{x+h})=(\color{orangered}{x+h})^2+4(\color{orangered}{x+h})+3\]

Answer 33

\[\large\rm \frac{f(x+h)-f(x)}{h}\]So we plug in our numerator,\[\large\rm \frac{(x+h)^2+4(x+h)+3-(x^2+4x+3)}{h}\]Distribute all the negatives and stuff, expand out the square,\[\large\rm \frac{x^2+2xh+h^2+4x+4h+3-x^2-4x-3}{h}\]Combine like-terms,\[\large\rm \frac{2xh+h^2+4h}{h}\]Divide out an h,\[\large\rm 2x+h+4\]Then the calculus part comes in,
this idea of "limits".
I dunno if you've been introduced to this?
You care about what happens to this function as h (the space between your two points in your slope formula) goes to zero.

What that means for us here,
is that we can directly plug 0 in for h.

Answer 34

\[\large\rm f'(x)=2x+4\]That gives us our derivative function,
this gives us the slope of the original function at any point.
We care about when the slope is 10,\[\large\rm 10=2x+4\]

Answer 35

Sorry if I took some of the fun away, doing the whole problem like that,
you seemed a bit confused though :d

Answer 36

thank you so much! so this equation basically defines all the locations on the graph, like if I plugged in a value for x and y?

Answer 37

Here is what our function looks like.

Answer 38

the derivative function tells us how the function is changing,
how steeply the function is growing at a particular moment in time.

So for example, here you can see that at x=-2,
the function is not growing at all, it's stationary,

while at x=1, it's very very steep, maybe a slope value of 5 or 6.