# Slope of a secant line

#### oregon88

Hey guys im having problems with this question

Consider the graph of the function f(x)=-5x^2+3x. Determine the slope of the secant line passing though the two points of the graph when x=1 and x=3. Determine the equation of the line passing through the two points.

#### dwsmith

MHF Hall of Honor
When x=1 and x=3, y=-2 and y=-36, respectively. How do you find a slope when given two points that form a line?

#### oregon88

You would use point slope form y-y1=m(x-x1) so can i just plug x=1 into the problem -5(1)^2+3x to get the solution?

#### dwsmith

MHF Hall of Honor
You would use point slope form y-y1=m(x-x1) so can i just plug x=1 into the problem -5(1)^2+3x to get the solution?
That is how you get the equation for the line. Did you ever get the value of m?

#### oregon88

I was working on it, to find the slope wouldnt you start with finding the
f '(x) (which from what I understand gives you the slope of the secant line and the slope of the tangent line)?

#### dwsmith

MHF Hall of Honor
I was working on it, to find the slope wouldnt you start with finding the
f '(x) (which from what I understand gives you the slope of the secant line and the of the tangent line)?
Slope equals rise over run. $$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$$

You have two points on your secant line so why derive?

#### oregon88

Ok so how do you find the slope of the secant line at those points x=1 and x=3 i guess is where im stuck. I understand at x=1 y=0 and at x=3 y=0.

#### dwsmith

MHF Hall of Honor
ok so how do you find the slope of the secant line at those points x=1 and x=3 i guess is where im stuck. I understand at x=1 y=0 and at x=3 y=0. what????
-5(1)^2+3(1)=-2

-5(3)^2+3(3)=-45+9=-36

#### oregon88

Oh so im making it harder than it really needs to be. So those are the slopes at those points. My last question is what does he mean by Determine the equation of the line passing through the two points.

#### dwsmith

MHF Hall of Honor
Oh so im making it harder than it really needs to be. So those are the slopes at those points. My last question is what does he mean by Determine the equation of the line passing through the two points.
Those are the y coordinates.

(1,-2) and (3,-36)

$$\displaystyle m=\frac{-36-(-2)}{3-1}=\frac{-34}{2}=-17$$

Find the equation of the line, use point-slope form.

$$\displaystyle y-y_1=m(x-x_1)\rightarrow y-(-2)=(-17)(x-1)\rightarrow y=-17x+15$$

oregon88