0
This is not a curve, but a line. It's slope will always be the same then, and it does not have tangent.
You can find it's rate of change easily enough though, that being it's slope. In this case, you can rewrite the equation as:
y = 13 - 2x
And you can see that it has a slope of negative two. This can be demonstrated as well by taking it's derivative:
y = 13 - 2 * x1
∴ y' = -2 * 1 * x1 - 1
∴ y' = -2 * 1 * x0
∴ y' = -2 * 1* 1
∴ y' = -2
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What is the value of k when y equals 3x plus 1 and is a line tangent to the curve of x squared plus y squared equals k?
k = 0.1
What is the value of y when y equals 2x plus 1.25 is a tangent to the curve y squared equals 10x?
If the line y = 2x+1.25 is a tangent to the curve y^2 = 10x then it works out that when x = 5/8 then y = 5/2
What is the value of k when the straight y equals 3x-1 is a tangent to the curve x squared plus y squared equals k?
2
What are the possible value of k when the line y equals x plus k is a tangent to the circle x squared plus y squared equals 25?
They are +/- 5*sqrt(2)
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
What is the value of k when y equals 3x plus 1 and is a line tangent to the curve of x squared plus y squared equals k?
k = 0.1
What is the value of k when y equals 3x plus 1 is a line tangent to the curve of x squared plus y squared equals k hence finding the point of contact of the line to the curve?
It is (-0.3, 0.1)
What is the value of y when y equals 2x plus 1.25 is a tangent to the curve y squared equals 10x?
If the line y = 2x+1.25 is a tangent to the curve y^2 = 10x then it works out that when x = 5/8 then y = 5/2
At what point is the line of y equals x -4 tangent to the curve of x squared plus y squared equals 8?
(2, -2)
What is the value of k in the line of y equals kx plus 1 and is tangent to the curve of y squared equals 8x?
If: y = kx+1 is a tangent to the curve y^2 = 8x Then k must equal 2 for the discriminant to equal zero when the given equations are merged together to equal zero.
What is the value of c when y equals x plus c is a tangent to the curve y equals 3 -x -5x squared hence finding the point of contact of the line with the curve showing work?
-2
What is the point of contact when the tangent line y equals x plus c touches the curve y equals 3x -x -5x squared?
-2
What is the equation to the tangent line to the curve xy plus y2 equals 0 at the point x equals 3 and y equals 0?
y=0. note. this is a very strange "curve". If y=0 then any value of x satisfies the equation, leading to a curve straight along the y axis. For any non-zero value of y the curve simplifies to y = -x. The curve is not differentiable at the origin.
How can you prove in at least two ways that the straight line of y equals 2x plus 5 over 4 is a tangent to the curve of y square equals 10x?
Combine the two equations together to give a quadratic equation in the form of:- 4x2 - 5x + 25/16 = 0 The solution to this is x = 5/8 or x = 5/8 meaning that it has equal roots therefore the line is tangent to the curve. The discriminant of b2 - 4ac = 0 also proves that the quadratic equation has two equal roots which makes the line tangent to the curve. Further proof can be found by plotting the straight line and curve graphically.
What is the value of k when the straight y equals 3x-1 is a tangent to the curve x squared plus y squared equals k?
2
How do you prove that the line y equals 2x plus 1.25 is tangent to the curve y squared equals 10x?
If the discriminant b2-4ac of the quadratic equation equals zero then it will have two equal roots meaning that the line is tagent to the curve. So by implication: (2x+1.25)(2x+1.25) = 10x 4x2-5x+25/16 = 0 Hence use the discriminant of b2-4ac :- (-5)2-4*4*25/16 = 0 Therefore the discriminant equals 0 so the line will be tangent to the curve. In fact working out the equation gives x having two equal roots of 5/8
How do you prove that the line y equals x-4 is tangent to the curve of x squared plus y squared equals 8?
equation 1: y = x-4 => y2 = x2-8x+16 when both sides are squared equation 2: x2+y2 = 8 Substitute equation 1 into equation 2: x2+x2-8x+16 = 8 => 2x2-8x+8 = 0 If the discriminant of the above quadratic equation is zero then this is proof that the line is tangent to the curve: The discriminant: b2-4ac = (-8)2-4*2*8 = 0 Therefore the discriminant is equal to zero thus proving that the line is tangent to the curve.
