Equations: y = kx +1.25 and y^2 = 10x
If: y = kx +1.25 then y^2 = (kx +1.25)^2 =>(kx)^2 +2.5kx +1.5625
So: (kx)^2 +2.5kx +1.5625 = 10x
Transposing terms: (kx)^2 +2.5kx +1.5625 -10x = 0
Using the discriminant formula: (2.5k -10)^2 -4(1.5625*k^2)
Multiplying out the brackets: 6.25k^2 -50k +100 -6.25^2 = 0
Collecting like terms: -50k +100 = 0
Solving the above equation: k = 2
Therefore the value of k is: 2
k = 2.
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.
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.
10 squared means 10*10 which equals 100.
If x squared = 5 then x = sqrt(5) = ± 2.2361 (approx).
The tangent ratio can take any real value.
k = 0.1
It is (-0.3, 0.1)
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
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.
2
-2
They are +/- 5*sqrt(2)
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.
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.
the shape of the curve skewed is "right"
14.2
By differentiating the answer and plugging in the x value along the curve, you are finding the exact slope of the curve at that point. In effect, this would be the slope of the tangent line, as a tangent line only intersects another at one point. To find the equation of a tangent line to a curve, use the point slope form (y-y1)=m(x-x1), m being the slope. Use the differential to find the slope and use the point on the curve to plug in for (x1, y1).