If: y = x^2 -8x +7 and y = kx -2
Then: X^2 -8x +7 = kx -2
Transposing terms: x^2 +(-8x -kx) +9 = 0
Using the discriminant: (-8 -k)^2 -4(1*9) = 0
Expanding brackets: 64 +16k +k^2 -36 = 0
Collecting like terms: k^2 +16k +28 = 0
Factorizing the above: (k +2)(k +14) = 0 meaning k = -2 or -14
Therefore the possible values of k are: -2 or -14
A tangent is a line that just touches a curve at a single point and its gradient equals the rate of change of the curve at that point.
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
To find the possible values of ( k ) for which the line ( y = kx - 2 ) is tangent to the curve ( y = x^2 - 8x + 7 ), we need to set the two equations equal and solve for ( x ): ( kx - 2 = x^2 - 8x + 7 ). Rearranging gives us the quadratic equation ( x^2 - (k + 8)x + 9 = 0 ). For the line to be tangent to the curve, this quadratic must have exactly one solution, which occurs when the discriminant is zero: ((k + 8)^2 - 4 \cdot 1 \cdot 9 = 0). Solving this gives ( k + 8 = \pm 6 ), leading to possible values ( k = -2 ) and ( k = -14 ).
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
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
A tangent is a line that just touches a curve at a single point and its gradient equals the rate of change of the curve at that point.
A line tangent to a curve, at a point, is the closest linear approximation to how the curve is "behaving" near that point. The tangent line is used to estimate values of the curve, near that point.
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
To find the possible values of ( k ) for which the line ( y = kx - 2 ) is tangent to the curve ( y = x^2 - 8x + 7 ), we need to set the two equations equal and solve for ( x ): ( kx - 2 = x^2 - 8x + 7 ). Rearranging gives us the quadratic equation ( x^2 - (k + 8)x + 9 = 0 ). For the line to be tangent to the curve, this quadratic must have exactly one solution, which occurs when the discriminant is zero: ((k + 8)^2 - 4 \cdot 1 \cdot 9 = 0). Solving this gives ( k + 8 = \pm 6 ), leading to possible values ( k = -2 ) and ( k = -14 ).
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
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
Gradient to the curve at any point is the derivative of y = x2 So the gradient is d/dx of x2 = 2x. When x = 2, 2x = 4 so the gradient of the tangent at x = 2 is 4.
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.
(2, -2)
You find the tangent to the curve at the point of interest and then find the slope of the tangent.
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.
It is (-0.3, 0.1)