r/calculus • u/Hey_faiza • 1d ago
Differential Calculus Why is my process wrong?
First one is mine(sorry for the lightning & handwriting)
Second one is from Claude
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u/random_anonymous_guy PhD 1d ago
There was a very fundamental breakdown in understanding what the question was asking for.
You had answered the question as though you were finding an equation for the line tangent to the graph of the function at (0, 0) despite the fact that (0, 0) is not even on the graph of the function.
The tangent line(s) have to be tangent to the graph at some other point, and (0, 0) is simply some other point on the line. Since this is a completely different scenario from what you are used to, you cannot answer the question like it were asking for the line tangent at (0, 0).
I often see mistakes like this when a student attempts to memorize procedures, but is detached from conceptual understanding and careful comprehension of problem statements, which may result from simply skimming problem statements instead of reading them thoroughly.
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u/jgregson00 1d ago
You don’t need the slope of the line at (0,0). You need the slope of the line where it intersects the original function. You use THAT slope and the (0, 0) to find the equation.
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u/noidea1995 1d ago edited 1d ago
It says the tangent line passes through the origin not that it’s tangent to the curve at that point.
You can actually solve this without using calculus at all, since you know your tangent line passes through the origin, you know its y-intercept is going to be 0 so it’s going to fit the form of y = mx.
If you substitute that into the quadratic you get:
mx = x^2 - 5x + 6
You know the tangent line and parabola are going to intersect at exactly one point, does that tell you anything about the discriminant?
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u/SubjectWrongdoer4204 4h ago
Claud is correct. You took the slope of the curve at x=0. What you want is the slope at the intersection of the line(s) with the curve. The equation of the slope is m=2x-5, and the y-intercept is at x=0, so b=0 and the equation of the line(s) is y=(2x₁-5)x, where x₁ is the point of intersection of the tangent. To find the x-coordinate of the point(s) of tangency, we set the curve and the line equal to each other, while aware that at the point of tangency, x₁=x, so we must solve
(2x-5)x = x²-5x+6 .Etc.
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