how do you know if the curve ever has a tangent line that has a slope which is an integer (like 5 or something)?
Isnt the tangent of a cubed function a squared function? is that even considered a tangent line? and how do you know the slope of it?
Isnt the tangent of a cubed function a squared function? is that even considered a tangent line? and how do you know the slope of it?
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If a function is differentiable, then every point on the graph of that function has a line tangent to the function at that point.
The EQUATION OF THE SLOPE OF ANY OF THOSE TANGENT LINES is f ' (x).
So, yes, you are correct that the EQUATION of the slope of the tangent line to ANY point on a cubic (3rd degree polynomial) is a quadratic (2nd degree polynomial).
To find a SPECIFIC slope, you need the x-coordinate of some point on the graph. You are usually given a point (x,y) and asked to find the tangent line. So you use m = f '(x), then substitute the value for x into f prime. That will be your slope m. Then use y = mx + b and substitute your point (x,y) to solve for b. Finally, write y = mx + b using the values for m and b.
----Oh, and Ray G is incorrect...f ' (x) is not guaranteed to be an integer.
The EQUATION OF THE SLOPE OF ANY OF THOSE TANGENT LINES is f ' (x).
So, yes, you are correct that the EQUATION of the slope of the tangent line to ANY point on a cubic (3rd degree polynomial) is a quadratic (2nd degree polynomial).
To find a SPECIFIC slope, you need the x-coordinate of some point on the graph. You are usually given a point (x,y) and asked to find the tangent line. So you use m = f '(x), then substitute the value for x into f prime. That will be your slope m. Then use y = mx + b and substitute your point (x,y) to solve for b. Finally, write y = mx + b using the values for m and b.
----Oh, and Ray G is incorrect...f ' (x) is not guaranteed to be an integer.
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Ok, so f(x) is your cubic function.
f'(x) is another function in x, and it gives you the slope of f(x) at the arbitrary point (x, f(x)).
Yes, it's a quadratic function.
Don't confuse f'(x) with the tangent line.
The line which is tangent at a given point (x0,f(x0)) has a constant slope = f(x0). That's just a single number. I've used x0 to clarify that this is just in reference to a specific point. The equation of the tangent line is y - f(x0) = f'(x0)(x-x0). That's just a linear function in x. Maybe it'll be clearer if we say
m0 = f'(x0) = slope at the tangent point
y0 = f(x0) = y-coordinate of a point on the cubic
and now
y - y0 = m0(x - x0) which should look more like a classic equation for a line.
There will be points where the slope is an integer. E.g., solve f'(x) = 5 to find the x-coordinates of the points where the cubic has slope 5. If the solutions are x=a and x=b, then the points where the cubic as slope 5 will be (a,f(a)) and (b, f(b). You have to plug a and b back into f to find the y-coordinates.
Hope all this helps!
f'(x) is another function in x, and it gives you the slope of f(x) at the arbitrary point (x, f(x)).
Yes, it's a quadratic function.
Don't confuse f'(x) with the tangent line.
The line which is tangent at a given point (x0,f(x0)) has a constant slope = f(x0). That's just a single number. I've used x0 to clarify that this is just in reference to a specific point. The equation of the tangent line is y - f(x0) = f'(x0)(x-x0). That's just a linear function in x. Maybe it'll be clearer if we say
m0 = f'(x0) = slope at the tangent point
y0 = f(x0) = y-coordinate of a point on the cubic
and now
y - y0 = m0(x - x0) which should look more like a classic equation for a line.
There will be points where the slope is an integer. E.g., solve f'(x) = 5 to find the x-coordinates of the points where the cubic has slope 5. If the solutions are x=a and x=b, then the points where the cubic as slope 5 will be (a,f(a)) and (b, f(b). You have to plug a and b back into f to find the y-coordinates.
Hope all this helps!
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Calculus is the easiest option.
If f(x) = ax^3 + bx^2 + cx + d
then the slope f'(x) = 3ax^2 + 2bx + c
There will always be x values - in fact an infinite number - where f'(x) is an integer.
If f(x) = ax^3 + bx^2 + cx + d
then the slope f'(x) = 3ax^2 + 2bx + c
There will always be x values - in fact an infinite number - where f'(x) is an integer.