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# What is tangent when x equals 4 and y equals -1?

Updated: 4/28/2022

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Q: What is tangent when x equals 4 and y equals -1?
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### What line is the tangent to y equals x squared when x equals 2?

To get the tangent, you first need to find it's slope: y = x2 &there4; 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: &Delta;y = s&Delta;x or: y - y1 = s(x - x1) &there4; y - 4 = 4(x - 2) &there4; y = 4 + 4x - 8 &there4; y = 4x - 4 And that is your tangent.

### What is the radius equation inside the circle x squared plus y squared -8x plus 4y equals 30 that meets the tangent line y equals x plus 4 on the Cartesian plane showing work?

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) =&gt; y = -x+2 Note that this proves that tangent of a circle is always at right angles to its radius

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### What is the gradient of the tangent to the curve at x equals 2 if Y equals x2?

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.

### What is the distance from a defined point on the x axis to the centre of circle x2 plus y2 -2x -6y plus 5 equals 0 when its tangent is at 3 4 on the Cartesian plane?

Equation of circle: x^2 +y^2 -2x -6y +5 = 0 Completing the squares: (x-1)^2 +(y-3)^2 = 5 Center of circle: (1, 3) Tangent contact point: (3, 4) Slope of radius: ((3-4)/(1-3) = 1/2 Slope of tangent line: -2 Equation of tangent line: y-4 = -2(x-3) =&gt; y = -2x+10 Equation tangent rearranged: 2x+y = 10 When y equals 0 then x = 5 or (5, 0) as a coordinate Distance from (5, 0) to (1, 3) = 5 using the distance formula

### If x2 plus y4 equals 10x plus 7 then what is the slope of the tangent line at 1 -2 and what is the equation of that tangent line- and at what point are the lines tangent to the original equation?

x^2+y^4=10x+7 @ (1,-2) Rearrange to isolate y. y=(-x^2+10x+7)^(1/4) Take derivative. y'=(1/4)*((-x^2+10x+7)^(-3/4))*(-2x+10) y'=(-2x+10)/4*(-x^2+10x+7)^(3/4) Sub in x=1. y'=8/2*(16)^(3/4) y'=2/8 y'=1/4 y'=m=slope=1/4 Sub m and (1,-2) into linear equation (tangent line). m*(x-x1)=y-y2 (1/4)(x-1)=y+2 Isolate y. Equation of tangent line: y=(1/4)x-(9/4). The third question is unclear.

### What are the equations of the tangent and normal at the coordinate of 6 4 of the circle x2 -4x plus y2 -6y equals 4 when plotted on the Cartesian plane?

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### What is the point of contact when the tangent line y equals x plus 4 meets the circle x squared plus y squared -8x plus 4y equals 30?

Circle equation: x^2 +y^2 -8x +4y = 30 Completing the squares: (x-4)^2 +(y+2)^2 = 50 Centre of circle: (4,-2) Slope of radius: -1 because it's perpendicular to the tangent line Radius equation: y--2 = -1(x-4) =&gt; y = -x+2 If: y = x+4 and y = -x+2 Then: x+4 =-x+2 or 2x = -2 =&gt; x = -1 Therefore by substitution the point of contact occurs at: (-1, 3)

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Assuming you mean the tangent at x = 4 of x&sup2; + y&sup2; = 26 then: x&sup2; + y&sup2; = 26 &rarr; y&sup2; = 26 - x&sup2; &rarr; y = &radic;(26 - x&sup2;) &rarr; slope = dy/dx = d/dx &radic;(26 - x&sup2;) = -x/&radic;(26 - x&sup2;) At x = 4: slope = -4/&radic;(26 - 4&sup2;) = -1.26491106.... &asymp; -1.265

### What is tangent 25 equals 7 divided by x?

x = 15.01, approx.