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How do you find a gradient of a linear equation if you are told that it is a tangent to another quadratic equation?
Wiki User
∙ 10y ago
- Solve the two equations simultaneously for x. You will probably need the y value as well.
- Differentiate the quadratic equation.
- Find the value of the derivative when you substitute the value for x (from step 1) into the derivative.
- That is the gradient.
- Solve the two equations simultaneously for x. You will probably need the y value as well.
- Differentiate the quadratic equation.
- Find the value of the derivative when you substitute the value for x (from step 1) into the derivative.
- That is the gradient.
- Solve the two equations simultaneously for x. You will probably need the y value as well.
- Differentiate the quadratic equation.
- Find the value of the derivative when you substitute the value for x (from step 1) into the derivative.
- That is the gradient.
- Solve the two equations simultaneously for x. You will probably need the y value as well.
- Differentiate the quadratic equation.
- Find the value of the derivative when you substitute the value for x (from step 1) into the derivative.
- That is the gradient.
Wiki User
∙ 10y ago
Wiki User
∙ 10y ago- Solve the two equations simultaneously for x. You will probably need the y value as well.
- Differentiate the quadratic equation.
- Find the value of the derivative when you substitute the value for x (from step 1) into the derivative.
- That is the gradient.
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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.
How to calculate the gradient from the non linear graph?
Draw a tangent to the curve at the point where you need the gradient and find the gradient of the line by using gradient = up divided by across
What is the definition of a tangent line?
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.
Does a quadratic has a even number of roots?
Yes, but only if you count a root at the tangent as a double root.
What are the possible values of k when y equals kx -2 which is tangent to the curve of y equals x squared -8x plus 7 showing work?
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
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.
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 does the rule for a quadratic function tell you about how the graph of the function will look?
If the quadratic function is written as ax2 + bx + c then if a > 0 the function is cup shaped and if a < 0 it is cap shaped. (if a = 0 it is not a quadratic) if b2 > 4ac then the equation crosses the x-axis twice. if b2 = 4ac then the equation touches the x-axis (is a tangent to it). if b2 < 4ac then the equation does not cross the x-axis.
What is the tangent equation for 2x3-3x2-8x plus 9 when x equals 2?
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
How to calculate the gradient from the non linear graph?
Draw a tangent to the curve at the point where you need the gradient and find the gradient of the line by using gradient = up divided by across
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.
How do you find the equation of a parabola if you know the equation of the tangent that touches it?
You need more than one tangent to find the equation of a parabola.
How do you find the equation of a tangent line?
In order to find the equation of a tangent line you must take the derivative of the original equation and then find the points that it passes through.
What is the definition of a tangent line?
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.
Derivative as the Slope of a Tangent?
Yes, the derivative of an equation is the slope of a line tangent to the graph.
Does a quadratic has a even number of roots?
Yes, but only if you count a root at the tangent as a double root.
How can you prove in at least two ways that the straight line of y equals 2x plus 5 over 4 is a tangent to the curve of y square equals 10x?
Combine the two equations together to give a quadratic equation in the form of:- 4x2 - 5x + 25/16 = 0 The solution to this is x = 5/8 or x = 5/8 meaning that it has equal roots therefore the line is tangent to the curve. The discriminant of b2 - 4ac = 0 also proves that the quadratic equation has two equal roots which makes the line tangent to the curve. Further proof can be found by plotting the straight line and curve graphically.
