To get the tangent, you first need to find it's slope:
y = x2
∴ dy/dx = 2x
So the slope with respect to x is equal to 2x. When x = 2 then, the slope is 4.
Now you need the y-coordinate that occurs when x equals 2. We know that y = x2, so when x equals 2, y equals 4.
You now have a point on the tangent, as well as it's slope, allowing you to define the line:
Recall the definition of a line:
Δy = sΔx
or:
y - y1 = s(x - x1)
∴ y - 4 = 4(x - 2)
∴ y = 4 + 4x - 8
∴ y = 4x - 4
And that is your tangent.
They are +/- 5*sqrt(2)
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
Circle equation: x^2 +y^2 -8x +4y = 30 Tangent line equation: y = x+4 Centre of circle: (4, -2) Slope of radius: -1 Radius equation: y--2 = -1(x-4) => y = -x+2 Note that this proves that tangent of a circle is always at right angles to its radius
That line is [ y = 2 cos(2x) ].
If: y = x-4 and y = x2+y2 = 8 then 2x2-8x+8 = 0 and the 3 ways of proof are: 1 Plot the given values on a graph and the line will touch the curve at one point 2 The discriminant of b2-4ac of 2x2-8x+8 must equal 0 3 Solving the equation gives x = 2 or x = 2 meaning the line is tangent to the curve
(2, -2)
They are +/- 5*sqrt(2)
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
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.
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.
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Circle equation: x^2 +y^2 -8x +4y = 30 Tangent line equation: y = x+4 Centre of circle: (4, -2) Slope of radius: -1 Radius equation: y--2 = -1(x-4) => y = -x+2 Note that this proves that tangent of a circle is always at right angles to its radius
That line is [ y = 2 cos(2x) ].
The question is suppose to read: Find the equation of the line tangent to y=(x²+3x)²(2x-2)³, when x=8
If: y = x-4 and y = x2+y2 = 8 then 2x2-8x+8 = 0 and the 3 ways of proof are: 1 Plot the given values on a graph and the line will touch the curve at one point 2 The discriminant of b2-4ac of 2x2-8x+8 must equal 0 3 Solving the equation gives x = 2 or x = 2 meaning the line is tangent to the curve
Circle equation: x^2 +y^2 -8x -16y -209 Completing the squares: (x-4)^2 +(y-8)^2 = 289 Centre of circle: (4, 8) Radius: 17 Slope of radius: 0 Tangent equation line: x = 21 passing through (21, 0)
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