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
What else can I help you with?
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 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 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 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
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 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.
What are the values of k when the line of y equals kx -2 is a tangent to the curve of y equals x squared -8x plus 7?
If: y = kx -2 and y = x^2 -8x+7 Then the values of k work out as -2 and -14 Note that the line makes contact with the curve in a positive direction or a negative direction depending on what value is used for k.
How can you prove that 2 equations ...x squared plus y squared equals 25... and a linear equation ...y equals -.75x plus 6.. are tangent to each other algebraically and without a calculator?
The solutions (x,y) to the first equation form the circle centered at 0, with radius 5 (the square root of 25). Solving for y, you can see that it consists of two functions of x: y = sqrt(25-x^2), y = -sqrt(25-x^2). Now you must show that y = -0.75x + 6 is tangent to one of them. That is, show that for one of the curves, at some point x', the slope is -0.75, AND the line intersects the curve at x' (that is, the y-values are equal for the curve and line, at point x'). Compute the derivative of each curve, and find for what x' it is equal to -0.75. (For each derivative, there is an easy integer solution which will work. No calculator needed.) Then check if the line intersects the curve at the point x' you find (for each curve). (It will work for one of the curves.) So you just showed that at point x', one of the curves has slope -0.75, and the line (also with slope -0.75) intersects the curve at x'. Therefore, the line is tangent to the curve at point x'. Hope that helps.
What are the values of k when kx plus y equals 4 is tangent to the curve y equals x squared plus 8 showing work?
If: kx+y = 4 and y = x^2+8 Then: x^2+8 = 4-kx => x^2+4+kx = 0 For the line to be tangent to the curve the discriminant of b^2-4(ac) must = 0 So: k^2-4(1*4) = 0 => k^2 -16 = 0 => k^2 = 16 => k = +/- 4 Therefore: y+4x = 4 and y-4x = 4 are tangents to the curve y = x^2+8
