Help with plotting equations pls

[AABatteries]

Hi!

I was just wondering if someone could guide me/give me tips on how to plot equations and know where to draw lines/parabolas.

I realised that since I haven't plotted in a while, that i've actually forgotten which is unfortunate.

This isn't for homework but just for work in class. Examples that I can think of at the top of my head are
- 3x+5
- 2x^2+5x-7
- X^2-12x+3
and ect. Thanks

 

[Deleted member 4993]

Hi!

I was just wondering if someone could guide me/give me tips on how to plot equations and know where to draw lines/parabolas.

I realised that since I haven't plotted in a while, that i've actually forgotten which is unfortunate.

This isn't for homework but just for work in class. Examples that I can think of at the top of my head are
- 3x+5
- 2x^2+5x-7
- X^2-12x+3
and ect. Thanks

A good primer would be:

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Please comeback and tell us if you fail to understand any point.

 

[HallsofIvy]

Lines are given by "linear equation" which have x to the first power only. The first of these equations, y*= 3x+ 5 (the "y= " is important!) has a straight line as graph. As we know from geometry "a line is 1.termined by two points". Taking x=0, y= 3(0)+ 5= 5 so one point on the line is (0, 5). Taking x= 1, y= 3(1)+ 5= 8. Another point on the line is (1, 8). The graph is the line through (0, 5) and (1, 8).



Polynomials with x to the second power ($\displaystyle x^2$) and no higher have graphs that are "parabolas". Parabolas are determined by 3 points and it is best if one of those points is the "vertex", You can determine the "vertex" by "completing the square".



The second problem has $\displaystyle y= 2x^2+5x-7$. I would start by factoring "2" out of the first two terms- $\displaystyle y= 2(x^2+ (7/2)x)+ 7$. I do that because I know that $\displaystyle (x+ a)^2= x^2+ 2ax+ a^2$. Comparing that to $\displaystyle x^2+ (7/2)x$ I see that I must have 2a= 7/2 so a= 7/4 and $\displaystyle a^2= 49/16$. I can "complete the square" by adding 49/16. Of course I don't want to change the actual value so I also subtract 49/16:

$\displaystyle y= 2x^2+ 5x- 7= 2(x^2+ (5/2)x+ 49/16- 49/16)- 7= 2(x+ 7/4)^2- 49/8- 7= 2(x+ 7/4)^2- 105/16$.



A square is never negative so the lowest point on the parabola is where that square is 0, at x= -7/4 and then y= -105/16. Further, the graph crosses the x-axis where $\displaystyle y= 2(x+ 7/4)^2- 105/16$, $\displaystyle 2(x+7/4)^2= 105/16$, $\displaystyle (x+ 7/4)^2= 105/32$, $\displaystyle x+7/4= \pm\sqrt{105/32}$, $\displaystyle x= -7/4\pm\sqrt{105/32}$. Just because it is easy, a fourth point is the "y-intercept" where x= 0 and $\displaystyle y= 2(7/4)^2- 105/16= 49/8- 105/16= 98/16- 105/16= -7/16$.



Plot those four points and, knowing the general shape of a "parabola", draw the graph through them



The third problem is easier because the coefficient of $\displaystyle x^2$ is 1 and the coefficient of $\displaystyle x$ is even.

$\displaystyle y= x^2- 12x+ 3= (x^2- 12x+ 36- 36)+3= (x- 6)^2- 33$.



The vertex is at (6, 33), the x-intercepts are at $\displaystyle (6+ \sqrt{33}, 0)$ and $\displaystyle (6- \sqrt{33}, 0)$ and the y=intercept is at $\displaystyle (0, 3)$.