Graphs, Linear and Quadratic Functions (Home Work)

Question

1. What is the equation of the line through (1,2) and (3, -1)?

2. What is the vertex of f(x) = x2 – 6x + 3?

3. Using points, straight lines, and possibly the curve of a parabola, construct a piecewise function whose graph looks like a smiling face.

Answer 1

(2)  In order to find the vertex of the parabola, we want to get the equation into the form:

y-b = (x-a)²

Then, (a,b) is the vertex.

So, to get there, we'll try and rewrite x²-6x+3 as (x-a)²

Well, (x-3)² would get us x²-6x+9

So let's change the equation from

y = x²-6x+3

to

y+6 = x²-6x+9 (by adding 6 to each side)

Now we have

y+6 = (x-3)²

and we can identify the vertex as (3,-6)

Answer 2

1) If you want to find the equation of a line, you should first find its slope.  m = (y2-y1)/(x2-x1), so (-1-2)/(3-1)=-3/2.  Now that you know the slope you can plug into the "point-slope" form of a line y-y1=m(x-x1).  This is actually the same equation as the slope formula, just change the y2 and x2 into y and x and multiply both sides by the (x2-x1).

 

so y-2=(-3/2)(x-1).

 

You can find the x value of the vertex by x = -b/2a = 6/2 = 3, then plug that in for x and find y = (3)²-6(3)+3 = -6.

 

 

Answer 3

1. Multiple lines can pass through the two point given. Did you want the linear equation of the line?

2. Vertex is at x = -b/2a, or x = -(-6)/2, x = 3.

3. Too silly for me to take my time.