I have the graph y = (x+1) (x+3)^2. I know that it is a cubic, but how do I distinguish whether it's a negative or positive one? Thanks.

Anyway, can someone please explain how once you have the points of x and y for the cubic graph, how you know how to sketch the curve (ie which way it goes)? Looking at this C1 solutionbank one below, …

the areas are both positive. for the last part i think it is looking for a new portion of the graph, this time above the x axis, which has the same area as R1 and R2.

Positive cubic graph. x=0 will be both a root and a turning point, similar to when you're sketching a graph of y=x Negative quadratic graph, with roots x=0 and x=6

As part of my maths work I have been asked 'Can you find a cubic function that has turning points at y = 1 and y = -1?' I am really stuck and starting to think it is impossible.

How do I show that a cubic equation only has on real root? Factorise it into where is a real root and is a quadratic which has discriminant less than 0.

Oct 26, 2024 · 0 = x^3 - 7 7 = x^3 ∛7 =x when y = 0 x = ∛7 and f (x) = (0)^3 - 7 f (x) = -7 when x = 0, y = -7 why does the curve for this equation look like it "slows" when approaching y = -7 I though that …

Jan 1, 2021 · Hi. When drawing cubic graphs like for example the question: - Sketch the curve y= (x-2) (2x-1) (2x+1) i know the shape of the curve, the roots and the y-int however I am confused about …

Jun 20, 2024 · Ignore previous post. The graph will touch the x-axis if 2 or 3 of the roots are the same. For example, the cubic graph (x+1) (x+2)^2 will cross the x-axis at -1 and touch the point -2, not go …

However it isn't possible to plot such a cubic in the classic 2D 'x versus f (x)' way. Obviously the only cubic you can plot in that way is one with real coefficients, and such a polynomial must have at least …