how do i graph this... it's basic, but i don't remember?

Question

i'm freshening up my mind with algebra 1 stuff that i don't remember exactly how to do, so that i can be ready for algebra 2...

how do i graph something like...

"graph the following"

1. y=3/5X +5


2. y=X2 -2X -2

(X2 means X squared)

thanks a bunch

Answer 1

1) you should know that the graph is a straight line
fill in two differnet x values, this gives you two y values, and givees you thus two different points (x1,y1) and (x2,y2)

2) this is a parabola, find the minimum/maximum by writing y = (x-1)^2 -3, then you see that for x = 1, the y value gets is minimum y =-3
next try to fimd the intersections with the x-axis, that is when y =0
thus solve 0=X2 -2X -2
if this has no solutions than you just choos two different x values calculate the y-values and draw a parabola througgh the 3 points.

CAPITO ??

Answer 2

1. I'm not sure whether you mean the linear equation

(a) y=(3/5)X+5 which can also be written y=3x/5 + 5

or

(b) y = 3/(5x) + 5 , which is inverse variation

The first (a) would be a simple linear graph. Join the y-intercept to the x intercept, and you have your line.

y-intercept (x=0): y = (3/5)0 + 5 = 5 , (0,5)
x-intercept (y=0): 0 = (3/5)x + 5
-3x/5 = 5, -3x = 25, x = -25/3 (-25/3,0)

Join the points and you have your line.

(b) If you are trying to graph the second equation it is a little more difficult to describe. To begin draw the line y=5 as a dotted line. The y-axis and the line y = 5 will be asymptotes for the graph of your function. The graph will approach but not cross these lines. Plot the following points: (10/3, 7) (5/3, 6) , 5/6, 11/2) and you will get the idea. Dont' forget graphing the negative portion of the graph.

2. This is the graph of a simple parabola. The y-intercept is (0,-2). Convert the equation to standard form for the co-ordinates of the vertex.

y = (x^2 - 2x + 1) - 3, y = (x-1)^2 - 3. The vertex is (1, -3)

If you want you can also determine the x-intecepts. The values are not going to work out as nicely as the values for the vertex and y-int.

y = (x-1)^2 - 3 or y = (x - 1 - sqrt3)(x - 1 + sqrt3)
sqrt = square root

So the x-intercepts are (1 + sqrt3,0) and (1 - sqrt3,0)

This should make graphing the parabola a piece of cake.

Answer 3

1. Straight Line

2. Parabola