Help with #8
@jim_thompson5910 Do you know how to do #8?
@surjithayer
I JUST DID THIS IN CLASS
Okay this is easy 1 second
Ok cool! :)
You know how to write it in standard form, yes?
Just write it in descending order of exponents.
Then the degree is 3. This is because it is the highest exponent.
Also, the degree is odd because it is 3. This means it has 3 solutions concluding that the end behavior must be different.
Ok so \[y=x^{3}+2x ^{2}-4x+8\]
That is correct, nice job
Ok, and the end behavior of the graph?
Well the degree is 3 right? This is because it's the highest exponent. The degree tells us how many SOLUTIONS the equation has. This means that the end behavior must be different. Either the graph will rise right, or fall left. We do not know since we are not given the leading coefficient.
So basically, if the degree is odd, the end behavior is always different. For an even degree, the end behavior would be the same.
Ok thanks! Did you already do this assignment?
No, I take algebra 2 at a public school.
Ah, ok. Do you think you could help me with a few more?
Yeah, of course.
Is it the same worksheet you linked?
Yeah, the next one is #14. I'll open a new question!
Ok, cool
\[y=x^3+2x^2-4x+8\] it is a polynomial of 3rd degree so three solutions. \[\frac{ dy }{ dx }=3x^2+4x-4\] \[\frac{ dy }{ dx }=0,~gives~3x^2+4x-4=0\] \[3x^2+6x-2x-4=0\] \[\left( x+2 \right)\left( 3x-2 \right)=0,x=-2,\frac{ 2 }{ 3 }\] tangent is parallel
tangent is parallel to x-axis at x=-2,2/3
@surjithayer She already got the answer.
\[\frac{ d^2y }{ dx^2 }=6x+4\] at x=-2 \[\frac{ d^2y }{ dx^2 }=-8<0\] hence there is maxima at x=-2 similarly it has minima at x=2/3 when x=-2 y=-8+8+8+8=16 whenx=2/3 \[y=\frac{ 8 }{ 27 }+\frac{ 8 }{ 9 }-\frac{ 8 }{ 3 }+8=\frac{ 8+24-72+216 }{ 27 }=\frac{ 176 }{ 27 }=6\frac{ 14 }{ 27 }\] now we can draw the graph |dw:1452130303643:dw|
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