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the line y=7x+58 goes through the point A. Prove that this line is a tangent to the circle at point a
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What circle?
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find the equation of a quadratic function whose graph is tangent at x=1 to the line whose slope8, tangent at x=-2 to solve the
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Start with defining the function: f(x)=ax²+bx+c Find its derivative: f'(x)=2ax+b 1. f'(1)=8 =>
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I have a two part question that pertains to a curve (r(x)) and its tangent line at x=3.
We are given that at x=3, r(x)=8. In
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I agree with your thinking for both questions. I especially like that you realize the relationship
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Find the equation of a TANGENT line & NORMAL line to the curve of x^2+y^2=20such that the tangent line is parallel to the line
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you want a tangent line with slope = 1/2 x^2+y^2 = 20 2x + 2yy' = 0 y' = -x/y For the tangent line,
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A tangent line is drawn to the hyerbola xy=c at a point P.
1) show that the midpoint of the line segment cut from the tangent
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To solve these problems, we'll use calculus and basic geometry. Let's go step by step: 1) To show
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please help, i procrastinated and now this is due tomorrow!!
A tangent line is drawn to the hyerbola xy=c at a point P. 1) show
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let p(a,c/a) be the point on the hyperbola for xy=c dy/dx = -y/x, so at P the slope = -c/a^2
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5. Let f be the function given by f(x) = x3- 7x + 6.
a. Find the zeros of f b. Write an equation of the line tangent to the graph
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Well, first off, if x=1 we have a zero. So divide by (x-1) and get x^2+x-6 factor that (x-2)(x+3) so
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Consider line segments which are tangent to a point on the right half (x>0) of the curve y = x^2 + 1 and connect the tangent
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I recall doing this question yesterday, it was a good one. I think you are stuck with expanding D^2
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Consider line segments which are tangent to a point on the right half (x>0) of the curve y = x^2+7 and connect the tangent point
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answer to C question at the end: If 1st coordinate (x-value) is C, the 2nd coordinate (y-value) is
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Consider line segments which are tangent to a point on the right half (x>0) of the curve y = x^2 + 1 and connect the tangent
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following the hints suggested: let the point be (c,c^2 + 1) dy/dx = 2x so at (c,c^2+1) the slope of
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for f(x)=lnx we know that f(e)=1. Use the tangent line at(e,1) to compute the tangent line approximation of f(3). what does the
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f'(x) = 1/x so, at (e,1), the slope is 1/e the line is thus (y-1)/(x-e) = 1/e y = 1/e (x-e) + 1 y =
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