Y=x^4+x^3-42x^2?

Question

what are its y and x intercepts and as x approaches infinity what does why approach positive or negative infinity and same with y

Answer 1

Factor out an x² to get
y=x²(x²+x-42) and then treat it just like the last one I explained to you. ☺

Doug

Answer 2

to get the x-int. you plug in 0 for Y because an x-int. is made when y=0.
so now you have 0=x^4+x^3-42x^2 and you need to factor it out.
x^2(x^2+x-42) =0
x^2(x+7)(x-6)=0 since anything times 0 is 0, all of these factors = 0.
so:
x^2=0
x=0

x+7=0
x=-7

x-6=0
x=6 so the x int. are: (0,-7), (0,0),(0,6)

The y-int.is when x=0. So plug in 0 for x.
y=0^4+0^3-42(0)^2.
y=0+0-0
y=0 so the y int. is: (0,0)

And the positive and negative infinity questions are asking what happens to x or y as you move positively or negatively.
So as x increases, does y increase or decrease?
As x decreases, does y increase or decrease?
Vice versa for y.
Graph the equation or plug in random coordinates in the equation to get your answers.