## mathcalculus Group Title Determine the slope of the secant line for the curve defined by the equation: (attached below) one year ago one year ago

1. mathcalculus Group Title

2. genius12 Group Title

The slope of the secant line is basically the average rate of change from $$\bf x_0$$ to $$\bf x_1$$. This is given by:$\bf m_{secant}=A.R.C.H|_{x_0}^{x_1}=\frac{ f(x_1)-f(x_0) }{ x_1-x_0 }$

3. genius12 Group Title

@mathcalculus Can you do that?

4. mathcalculus Group Title

is that like the differentiation but in other terms?

5. mathcalculus Group Title

@genius12

6. mathcalculus Group Title

is it = to 1?

7. mathcalculus Group Title

do we substitute just for it?

8. mathcalculus Group Title

can we solve this together? @genius12

9. mathcalculus Group Title

im stuck.

10. genius12 Group Title

Ok I will solve step by step. Firstly, you should realise that this is essentially the slope formula which is:$\bf m=\frac{ y_2-y_1 }{ x_2-x_1 }$You remember that formula? @mathstudent55

11. mathcalculus Group Title

yes

12. mathcalculus Group Title

that's the formula to find the slope .

13. mathcalculus Group Title

are you there?

14. genius12 Group Title

Well that what this essentially is! It's the slope, of the secant line, and the secant line is the line that connects the two points on the graph when x = -5 to when x = -6. I'll draw the graph and show to you:|dw:1375078386992:dw|

15. mathcalculus Group Title

well i used that formula you gave me and i got 1/1

16. genius12 Group Title

the y2 and y1 are such the y values at the -5 and -6. So plug in x = -5 and x = -6 in to f(x) = -2x^2 - 1 and find the y-values of each. Now just calculate the slope with the:$\bf m=\frac{y_2-y_1}{x_2-x_1}=\frac{ f(-6)-f(-5) }{ -6-(-5) }$

17. genius12 Group Title

$\bf f(-6)=-2(-6)^2-1=-73$$\bf f(-5)=-2(-5)^2-1=-51$Plugging these values in we get:$\bf slope_{secant}=\frac{ f(-6)-f(-5) }{ -6-(-5) }=\frac{ -73-(-51) }{ -6-(-5) }=\frac{ -22 }{ -1 }=22$

18. genius12 Group Title

@mathcalculus That's what you are supposed to do. How did you get 1?

19. mathcalculus Group Title

so it;s 22?

20. genius12 Group Title

Yes. How did you get 1?

21. mathcalculus Group Title

im not sure i did it the other way around.

22. mathcalculus Group Title

hey, thank you so much , i'll be here tomorrow. need to sleep, thanks again. @genius12

23. genius12 Group Title

Well I'll list out the steps for you. Find the y-value at both x-values. Here we were finding the slope of the secant line (the line that connects the two y-values at x = -5 and x = -6). We first find the y-value at x = -6 by plugging -6 for x in to f(x). Then we find y-value at x = -5 by plugging in -5 for x. When we have these 2 y-values, you just use the slope formula.

24. mathcalculus Group Title

lol thank you ^_^