Cubic+ Quad Equations - y Axis

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The discussion centers on the cubic equation x³ - 7x + 1 = 0, derived from rearranging x³ - 5x = 2x - 1. The y-intercept of the function y = x³ - 7x + 1 is confirmed to be 1, occurring when x = 0. The term "gradient" is clarified as not applicable to cubic equations in the same manner as linear equations, as cubic functions do not possess a constant slope. Instead, the concept of a derivative may be relevant for understanding the rate of change of the function.

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thomas49th
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Hi, I was wondering in a cubic equation such as

x³ - 5x = 2x - 1
can be rearranged to

x³ - 7x + 1 = 0

is that 0 the y value, and +1 where the line crosses the y axis? Is - 7x the gradient or is x³ - 7x the gradient?

Thankyou
 
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"is that 0 the y value"?? What y value? An equation is not a function. If you are talking about y= x3- 7x+ 1, then the "y-intercept" is the value of y when x= 0: y= 1- the y coordinate when the curve (not a straight line) crosses the y-axis. Of course, we still have a value of y for every value of x. I'm not clear what you mean by "the gradient". Since this is NOT a straight line, it doesn't have a "slope" (sometimes called "gradient"). Sometimes the term "gradient" is used to mean "derivative" but, from the problems you have been posting here I would not expect you to be referring to that.
 
erm...

Question 14 of a paper is:

A graph -2< x < 3 with points

x = -2, -1 , 0 , 1 , 2 ,3
y = 1 , 4 , 1 ,-2 , 1 ,16

I drew the graph. Now part b says

By drawing suitable lines on you graph, find all the solutions to the following equations for -2 < x < 3. Give your solutions to 1 decimal place

x³ - 5x = 2x - 1

so x³ - 7x + 1 = 0

dont know where to go from there...
 

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