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Find the equation of the tangent line at the given value y equals ln x squared at x equals 1?
Given y=LN(x2). To find the equation of a tangent line to a point on a graph, we must figure out what the slope is at that exact point. We must take the derivitave with respect to x to come up with a function that will give the slope of the tangent line at any point on the graph. Any text book will tell you that the derivative with respect to X of LN(U) is the quantity U prime over U(In other words, the derivitive of U over U. U is whatever is inside the natural log). In this case, it is y = (2x/x2). To find the slope, we then plug in an x coordinate, 1. We get y=2. The slope of the tangent line at this point is 2. We now have m for the slope-intercept form y = mx + b. Now we must find b. B is the y intercept, aka where does the graph intersect the y axis. We use the slope. We know that the graph of the tangent line contains the point (0,1). We go down 2 and left one, to find that the tangent line intercepts the y axis at -1. We now have B. The equation of the tangent line of y = LN(x2) at X = 1 is y = 2X-1. Hope this helps!!!
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Where is the center of the circle given by the equation x - 3 squared plus y plus 2 squared equals 9?
The center of the circle given by the equation (x - 3)2 plus (y + 2)2 = 9 is (3,-2).
How do you determine if a given expression is an equation?
An equation has an equals sign ( = ). Equations assert the absolute equality of two expressions.
What is the slope of the line given by the equation y equals 11.2x?
Well, to answer your question,
What is the solution set of the equation x squared plus 3x - 46?
49
What is the solution for m given the equation 4m - 29 equals -105?
m = -19
How do you find the length of a leg on a right triangle when one leg and the slope of the hypotenuse is given?
Use tangent. Your equation will be tan(slope of hypotenuse) = opposite side / adjacent side. it's easier if you just do A squared plus b squared equals c squared. Then subtitute the numbers gived in.
How do you get the tangent line without the graph?
The answer will depend on the context. If the curve in question is a differentiable function then the gradient of the tangent is given by the derivative of the function. The gradient of the tangent at a given point can be evaluated by substituting the coordinate of the point and the equation of the tangent, though that point, is then given by the point-slope equation.
Where is the center of the circle given by the equation x plus 4 squared plus y plus 6 squared equals 49?
(-4,-6)
Where is the center of the circle given by the equation x plus 5 squared plus y minus 3 squared equals 25?
The centre is (-5, 3)
Where is the center of the circle given by the equation x - 3 squared plus y plus 2 squared equals 9?
The center of the circle given by the equation (x - 3)2 plus (y + 2)2 = 9 is (3,-2).
What is the value of k in the line of y equals kx plus 1 and is tangent to the curve of y squared equals 8x?
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.
In which direction does the parabola that is given by the equation below open..... x equals -2.5 parenthese y - 3 end parenthese squared plus 5?
Left
How do you calculate the tangent on a distance time graph?
You can calculate the tangent for a give time, T, as follows: Substitute the value of the time in the distance-time equation to find the distance at the given time. Suppose it is f(T). Differentiate the distance-time equation with respect to time. For any given time, substitute its value in the derivative and evaluate. That is the gradient of the tangent, v. Then equation of the tangent is f(T) - f(t) = v*(T - t)
What is the relationship if any between the curves of y equals x squared -4x plus 8 and y equals 8x -14 -x squared with explanations?
There is no connection between the given curves because when they are combined into a single quadratic equation the discriminant of the equation is less than zero which means they share no valid roots.
How do you prove in 3 different ways that the line of y equals x -4 is tangent to the curve of x squared plus y squared equals 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
How do you prove that the straight line y equals x plus c is a tangent to the curve x squared plus y squared equals 4 when c squared equals 8?
The general form of a line tangent to a circle is:y=mx+a(1+m2 )1/2. where "a" is the radius of the circle. Here circle is x2 + y2=4, so radius=a=2. nowc2=a2(1+m2)=8 (given)or 8=4(1+m2)2=1+m2 orm2=1 orm=1. so equation becomesy=mx+c ory=x+cImproved Answer:Equation 1: y = x+square root of 8 => y2 = x2+square root of 32x+8 when both sides are squared.Equation 2: y2 = 4-x2By definition:x2+the square root of 32x+8 = 4-x2 => 2x2+the square root of 32x+4 = 0If the discriminant b2-4ac of the above quadratic equation is equal to zero then this is proof that the straight line is tangent to the curve:b2-4ac = the square root of 322-4*2*4 = 0Therefore the straight line is a tangent to the curve because the discriminant of the quadratic equation equals zero.
What is the given points for the equation y equals 2x-3?
5
