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Describe the curve x2 plus y2 equals 25?
A circle, centre (0,0), radius = 5
What are the possible values of k in the line y equals kx -2 which is tangent to the curve y equals x squared -8x plus 7?
The possible values for k are -2 and -14 because in order for the line to be tangent to the curve the discriminant must be equal to 0 as follows:- -2x-2 = x2-8x+7 => 6-x2-9 = 0 -14x-2 = x2-8x+7 => -6-x2-9 = 0 Discriminant: 62-4*-1*-9 = 0
How do you find the exact value of the gradient of the curve with equation y equals 1 divide by 4 minus x2 at the point where x equals 1 over 2?
You find the gradient of the curve using differentiation. The answer is 0.07111... (repeating).
What does the Graph of x2 25 look like?
If f(x) = x2 + 25, then to plot f(x) on a graph would give you a parabolic curve extending infinitely upward with a minimum value of 25, and it's vertex at the point (0, 25).
What are the possible values of k when y equals kx -2 which is tangent to the curve of y equals x squared -8x plus 7 showing work?
The gradient to the curve y = x2 - 8x + 7 is dy/dx = 2x - 8The gradient of the tangent to the curve is, therefore, 2x - 8.The gradient of the given line is kTherefore k = 2x - 8. That is, k can have ANY value whatsoever.Another Answer:-If: y = kx-2 and y = x2-8x+7Then: x2-8x+7 = kx-2 => x2-8x-kx+9 = 0Use the discriminant of: b2-4ac = 0So: (-8-k)2-4*1*9 = 0Which is: (-8-k)(-8-k)-36 = 0 => k2+16k+28 = 0Using the quadratic equation formula: k = -2 or k = -14 which are the possible values of k for the straight line to be tangent with the curve
