bvruh how do i sketch (1 Viewer)

[english? ensquish]

hey can someone help me out in sketching y=x^4-8x^2+11 and y=3x^3-x+2

for the first one i tried finding a factor first to long divide which didn't work and then tried using a sub but also couldn't get the intercepts.
and second one i got 3 intercepts through long division when there's only supposed to be 1.

any help would be appreciated thanks

 

[mmmmmmmmaaaaaaa]

Firstly, you can test for when x--> infinity and x-->-infinity
For the first one, by doing so, you will find that they approach +infinity for both. ie. go up on both sides.
You will also find that the y-intercept is 11.
To find the x intercepts, use the quadratic formula where x^2=(quad formula).
And to find the turning points, use first derivative

 

[chilli 412]

for the first one:

 

[chilli 412]

for the second one

remember for long division to always have your dividend in the general polynomial form (such as ax^3 + bx^2 +cx + d for cubics) which is why i wrote 0x^2 in my dividend

 

[english? ensquish]

for the second one
View attachment 36751
remember for long division to always have your dividend in the general polynomial form (such as ax^3 + bx^2 +cx + d for cubics) which is why i wrote 0x^2 in my dividend

thank you so much, i have a question on how you determined the 4 possible roots for f(x)=0.

 

[chilli 412]

thank you so much, i have a question on how you determined the 4 possible roots for f(x)=0.

for f(x) = 0, we have the equation x^4 -8x^2 +11 = 0
we can notice that x^4 is the same as (x^2)^2 by power laws
so we actually have a hidden quadratic in this equation of degree 4, and if we replace x^2 with 'u' , this means that x^4 = u^2 , and 8x^2 is 8u. we can then use the quadratic formula to find 2 possible roots, but because these two roots are solutions for x^2, this means we must take the plus or minus (±) square root of our solutions. this then leaves us with four possible solutions, hope this helped