# Math Help - Sketching graphs, dom and ran, quartics

1. ## Sketching graphs, dom and ran, quartics

Hello,
I needed some help in solving the following:
Sketch the graphs with the following equations, showing any axial intercepts. State the maximal domain and range
Q1.
(Sorry for not knowing how to use texify)
Q2. y= 1
--------- -4
(x+1)^2

Q3. Write down an equation, in factor form, for the graph show below:

(can't draw it, but it's a quartic function which touches the x-axis at -2 and 1. The y-intercept is 20, however this is not the middle of the graph, the graph appears to have been shifted 1 unit to the left, i hope it isn't too hard to picture what im trying to say.

I honestly have no clue on how to do the first two, the third i attempted
because i no the equation is supposed to be something like: y=(x+2)(x-1)
but i'm not sure on how to obtain y-intercepts. please provide a detailed explanation so i can understand this, thank you math forum

2. Originally Posted by 99.95
Hello,
I needed some help in solving the following:
Sketch the graphs with the following equations, showing any axial intercepts. State the maximal domain and range
Q1.
(Sorry for not knowing how to use texify)
I have no idea what "texify" is. You can use "LaTex" on this board. There is a tutorial in the "Math Resources" section.

Q2. y= 1
--------- -4
(x+1)^2
Obviously x cannot be -1 because then the denominator would be 0. a fraction is only 0 when the numerator is 0 and here the numerator, 1, is never 0. The fraction is never 0 which means y is never -4.

Q3. Write down an equation, in factor form, for the graph show below:

(can't draw it, but it's a quartic function which touches the x-axis at -2 and 1.

The y-intercept is 20, however this is not the middle of the graph, the graph appears to have been shifted 1 unit to the left, i hope it isn't too hard to picture what im trying to say.
Saying that it is a quartic means it of fourth degree and can be written y= a(x-b)(x-c)(x-d)(x-e). You say it "touches" the x-axis which I am going to assume means it does not cross the axis but is tangent to the x-axis there. That means that -2 and 1 are double roots and $y= a(x-(-2))(x-(-2))(x- 1)(x- 1)= a(x+2)^2(x- 1)^2$. The fact that the y-intercept is 20 means that $y(0)= a(0+ 2)^2(0- 1)^2= 4a= 20$ so a= 5.

[/quote]I honestly have no clue on how to do the first two, the third i attempted
because i no the equation is supposed to be something like: y=(x+2)(x-1)[/quote]
If you mean the curve was quadratic, not "quartic", then, yes, y= a(x+2)(x-1). The "y-intercept" is the y value when x= 0. Set x=0, y= 20 and solve for a.

but i'm not sure on how to obtain y-intercepts. please provide a detailed explanation so i can understand this, thank you math forum