The tangent to 2x3 - 3x2 - 8x + 9 at x = 2 is y = 4x - 11
The tangent to y = 2x3 - 3x2 - 8x + 9 at x = 2 has the same gradient as the curve at that point; to find the gradient, differentiate:
dy/dx = 6x2 - 6x - 8
which at x = 2 is:
gradient = 6 x 22 - 6 x 2 - 8 = 4
At x = 2, y = 2 x 23 - 3 x 22 - 8 x 2 + 9 = -3
The equation of a line through point (xo, yo) with gradient m is:
y - yo = m(x - xo)
Thus the equation of the tangent to the line at x = 2 is:
y - -3 = 4(x - 2)
⇒ y = 4x - 11
-2
A linear equation.
It is an equation of a straight line.
Equation of circle: x^2 +y^2 -6x +4y +5 = 0 Completing the squares: (x-3)^2 +(y+2)^2 = 8 Radius of circle: square root of 8 Center of circle: (3, -2) Circle makes contact with the x axis at: (1, 0) and (5, 0) Slope of 1st tangent: 1 Slope of 2nd tangent: -1 1st tangent line equation: y = 1(x-1) => y = x-1 2nd tangent line equation: y = -1(x-5) => y = -x+5
3d + 14 = 11 is the equation.
(a) y = -3x + 1
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
Equation of circle: x^2 +10x +y^2 -2y -39 = 0 Completing the squares: (x+5)^2 +(y-1)^2 = 65 Center of circle: (-5, 1) Point of contact: (3, 2) Slope of radius: 1/8 Slope of tangent: -8 Tangent equation: y-2 = -8(x-3) => y = -8x+26
Circle equation: x^2 +y^2 +6x -10y = 0 Completing the squares: (x +3)^2 +(y -5)^2 = 34 Center of circle: (-3, 5) Point of contact: (0, 0) Slope of radius: -5/3 Slope of tangent line: 3/5 Tangent line equation: y = 0.6x
A quadratic equation.
Iron plus chlorine equals Iron chloride is the word equation.
2
That's the equation.
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.
x = 0 and y = 4
k = 0.1
The tangent equation that touches the circle 2x^2 +2y^2 -8x -5y -1 = 0 at the point of (1, -1) works out in its general form as: 4x +9y +5 = 0